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Question: A solution contains A$^\oplus$ and B$^\oplus$ in such a concentration that both deposits simultaneou...

A solution contains A⊕^\oplus and B⊕^\oplus in such a concentration that both deposits simultaneously. If current of 9.65 amp was passed through 100 ml solution for 55 seconds then find the final concentration of A⊕^\oplus ions if initial concentration of B⊕^\oplus is 0.1M.

Given : A⊕^\oplus + e−^- ⟶\longrightarrow A; E° = -0.5 volt

B⊕^\oplus + e−^- ⟶\longrightarrow B; E° = -0.56 volt

2.303RTF\frac{2.303RT}{F} = 0.06

Answer

0.005 M

Explanation

Solution

To solve this problem, we need to consider the conditions for simultaneous deposition and the effect of passing an electric current.

1. Determine the initial concentration of A⊕^\oplus for simultaneous deposition: When two ions deposit simultaneously, their reduction potentials must be equal. We use the Nernst equation: E=E∘−2.303RTnFlog⁡1[ion]E = E^\circ - \frac{2.303RT}{nF} \log \frac{1}{[\text{ion}]} Given 2.303RTF=0.06\frac{2.303RT}{F} = 0.06 and n=1n=1 for both ions, the equation simplifies to: E=E∘+0.06log⁡[ion]E = E^\circ + 0.06 \log [\text{ion}]

For simultaneous deposition: EA⊕/A=EB⊕/BE_{A^\oplus/A} = E_{B^\oplus/B} EA⊕/A∘+0.06log⁡[A⊕]initial=EB⊕/B∘+0.06log⁡[B⊕]initialE^\circ_{A^\oplus/A} + 0.06 \log [A^\oplus]_{initial} = E^\circ_{B^\oplus/B} + 0.06 \log [B^\oplus]_{initial}

Given: EA⊕/A∘=−0.5E^\circ_{A^\oplus/A} = -0.5 V EB⊕/B∘=−0.56E^\circ_{B^\oplus/B} = -0.56 V [B⊕]initial=0.1[B^\oplus]_{initial} = 0.1 M

Substitute the values: −0.5+0.06log⁡[A⊕]initial=−0.56+0.06log⁡(0.1)-0.5 + 0.06 \log [A^\oplus]_{initial} = -0.56 + 0.06 \log (0.1) −0.5+0.06log⁡[A⊕]initial=−0.56+0.06×(−1)-0.5 + 0.06 \log [A^\oplus]_{initial} = -0.56 + 0.06 \times (-1) −0.5+0.06log⁡[A⊕]initial=−0.56−0.06-0.5 + 0.06 \log [A^\oplus]_{initial} = -0.56 - 0.06 −0.5+0.06log⁡[A⊕]initial=−0.62-0.5 + 0.06 \log [A^\oplus]_{initial} = -0.62 0.06log⁡[A⊕]initial=−0.62+0.50.06 \log [A^\oplus]_{initial} = -0.62 + 0.5 0.06log⁡[A⊕]initial=−0.120.06 \log [A^\oplus]_{initial} = -0.12 log⁡[A⊕]initial=−0.120.06=−2\log [A^\oplus]_{initial} = \frac{-0.12}{0.06} = -2 [A⊕]initial=10−2 M=0.01 M[A^\oplus]_{initial} = 10^{-2} \text{ M} = 0.01 \text{ M}

2. Determine the ratio of concentrations for simultaneous deposition: As long as both ions deposit simultaneously, their reduction potentials remain equal. EA⊕/A∘+0.06log⁡[A⊕]=EB⊕/B∘+0.06log⁡[B⊕]E^\circ_{A^\oplus/A} + 0.06 \log [A^\oplus] = E^\circ_{B^\oplus/B} + 0.06 \log [B^\oplus] 0.06(log⁡[A⊕]−log⁡[B⊕])=EB⊕/B∘−EA⊕/A∘0.06 (\log [A^\oplus] - \log [B^\oplus]) = E^\circ_{B^\oplus/B} - E^\circ_{A^\oplus/A} 0.06log⁡([A⊕][B⊕])=−0.56−(−0.5)0.06 \log \left(\frac{[A^\oplus]}{[B^\oplus]}\right) = -0.56 - (-0.5) 0.06log⁡([A⊕][B⊕])=−0.060.06 \log \left(\frac{[A^\oplus]}{[B^\oplus]}\right) = -0.06 log⁡([A⊕][B⊕])=−1\log \left(\frac{[A^\oplus]}{[B^\oplus]}\right) = -1 [A⊕][B⊕]=10−1=0.1\frac{[A^\oplus]}{[B^\oplus]} = 10^{-1} = 0.1 This ratio must be maintained throughout the simultaneous deposition process.

3. Calculate the total moles of electrons passed: Current I=9.65I = 9.65 A Time t=55t = 55 s Charge Q=I×t=9.65 A×55 s=530.75 CQ = I \times t = 9.65 \text{ A} \times 55 \text{ s} = 530.75 \text{ C} Moles of electrons ne=QFn_e = \frac{Q}{F}, where Faraday's constant F=96500 C/molF = 96500 \text{ C/mol}. ne=530.75 C96500 C/mol=0.0055 moln_e = \frac{530.75 \text{ C}}{96500 \text{ C/mol}} = 0.0055 \text{ mol}

4. Calculate the initial moles of A⊕^\oplus and B⊕^\oplus: Volume of solution V=100 ml=0.1 LV = 100 \text{ ml} = 0.1 \text{ L} Initial moles of A⊕^\oplus: nA,initial=[A⊕]initial×V=0.01 M×0.1 L=0.001 moln_{A,initial} = [A^\oplus]_{initial} \times V = 0.01 \text{ M} \times 0.1 \text{ L} = 0.001 \text{ mol} Initial moles of B⊕^\oplus: nB,initial=[B⊕]initial×V=0.1 M×0.1 L=0.01 moln_{B,initial} = [B^\oplus]_{initial} \times V = 0.1 \text{ M} \times 0.1 \text{ L} = 0.01 \text{ mol}

5. Calculate the moles of A⊕^\oplus and B⊕^\oplus deposited: Let ΔnA\Delta n_A be the moles of A⊕^\oplus deposited and ΔnB\Delta n_B be the moles of B⊕^\oplus deposited. Since both ions are monovalent, the total moles of ions deposited equals the moles of electrons passed: ΔnA+ΔnB=ne=0.0055 mol\Delta n_A + \Delta n_B = n_e = 0.0055 \text{ mol}

The final concentrations must also satisfy the simultaneous deposition ratio: [A⊕]final[B⊕]final=0.1\frac{[A^\oplus]_{final}}{[B^\oplus]_{final}} = 0.1 Since the volume is constant, this implies: nA,finalnB,final=0.1\frac{n_{A,final}}{n_{B,final}} = 0.1 nA,final=0.1×nB,finaln_{A,final} = 0.1 \times n_{B,final} (nA,initial−ΔnA)=0.1×(nB,initial−ΔnB)(n_{A,initial} - \Delta n_A) = 0.1 \times (n_{B,initial} - \Delta n_B) 0.001−ΔnA=0.1×(0.01−ΔnB)0.001 - \Delta n_A = 0.1 \times (0.01 - \Delta n_B) 0.001−ΔnA=0.001−0.1ΔnB0.001 - \Delta n_A = 0.001 - 0.1 \Delta n_B ΔnA=0.1ΔnB\Delta n_A = 0.1 \Delta n_B

Now we have a system of two equations:

  1. ΔnA+ΔnB=0.0055\Delta n_A + \Delta n_B = 0.0055
  2. ΔnA=0.1ΔnB\Delta n_A = 0.1 \Delta n_B

Substitute (2) into (1): 0.1ΔnB+ΔnB=0.00550.1 \Delta n_B + \Delta n_B = 0.0055 1.1ΔnB=0.00551.1 \Delta n_B = 0.0055 ΔnB=0.00551.1=0.005 mol\Delta n_B = \frac{0.0055}{1.1} = 0.005 \text{ mol}

Now find ΔnA\Delta n_A: ΔnA=0.1×0.005=0.0005 mol\Delta n_A = 0.1 \times 0.005 = 0.0005 \text{ mol}

6. Calculate the final concentration of A⊕^\oplus ions: nA,final=nA,initial−ΔnA=0.001 mol−0.0005 mol=0.0005 moln_{A,final} = n_{A,initial} - \Delta n_A = 0.001 \text{ mol} - 0.0005 \text{ mol} = 0.0005 \text{ mol} [A⊕]final=nA,finalV=0.0005 mol0.1 L=0.005 M[A^\oplus]_{final} = \frac{n_{A,final}}{V} = \frac{0.0005 \text{ mol}}{0.1 \text{ L}} = 0.005 \text{ M}

The final concentration of A⊕^\oplus ions is 0.005 M.