Question

Given that E°(Zn$^{2+}$/Zn) = -0.764 V and E°(Cd$^{2+}$/Cd) = -0.402 V, the emf of the following cell: Zn | Zn$^{2+}$ (0.2 M) || Cd$^{2+}$ (0.004 M) | Cd is given by

Answer 1

0.312 V

Answer 2

The given electrochemical cell is Zn | Zn$^{2+}$ (0.2 M) || Cd$^{2+}$ (0.004 M) | Cd.

The standard electrode potentials are: E°(Zn$^{2+}$/Zn) = -0.764 V E°(Cd$^{2+}$/Cd) = -0.402 V

1. Determine the anode and cathode:

The species with the more negative standard reduction potential undergoes oxidation (anode). Comparing -0.764 V (Zn) and -0.402 V (Cd), Zn has a more negative potential. Therefore, Zn acts as the anode and Cd acts as the cathode.

2. Write the half-cell reactions and the overall cell reaction:

Anode (Oxidation): Zn(s) → Zn$^{2+}$(aq) + 2e$^-$ Cathode (Reduction): Cd$^{2+}$(aq) + 2e$^-$ → Cd(s) Overall cell reaction: Zn(s) + Cd$^{2+}$(aq) → Zn$^{2+}$(aq) + Cd(s) From the overall reaction, the number of electrons transferred (n) is 2.

3. Calculate the standard cell potential (E°_cell):

E°_cell = E°_cathode - E°_anode E°_cell = E°(Cd$^{2+}$/Cd) - E°(Zn$^{2+}$/Zn) E°_cell = (-0.402 V) - (-0.764 V) E°_cell = -0.402 V + 0.764 V E°_cell = 0.362 V

4. Calculate the reaction quotient (Q):

For the reaction Zn(s) + Cd$^{2+}$(aq) → Zn$^{2+}$(aq) + Cd(s), the reaction quotient is: $Q = \frac{[Zn^{2+}]}{[Cd^{2+}]}$

Given concentrations: [Zn$^{2+}$] = 0.2 M and [Cd$^{2+}$] = 0.004 M $Q = \frac{0.2}{0.004} = \frac{200}{4} = 50$

5. Apply the Nernst equation to find the cell emf (E_cell):

The Nernst equation at 25°C is: $E_{cell} = E^\circ_{cell} - \frac{0.0592}{n} \log_{10} Q$

Substitute the calculated values: $E_{cell} = 0.362 - \frac{0.0592}{2} \log_{10} (50)$ $E_{cell} = 0.362 - 0.0296 \times \log_{10} (50)$

We know that $\log_{10} (50) \approx 1.699$

$E_{cell} = 0.362 - 0.0296 \times 1.699$ $E_{cell} = 0.362 - 0.0503$ $E_{cell} = 0.3117$ V

Rounding to three decimal places, the emf of the cell is 0.312 V.