Solveeit Logo
Login

Question

Question: Consider the following electrodes A) $Cu^{2+}_{(aq)} | Cu_{(s)}$ $10^{-4} m$ B) $Cu^{2+}_{(aq)} | Cu...

Consider the following electrodes A) Cu(aq)2+∣Cu(s)Cu^{2+}_{(aq)} | Cu_{(s)} 10−4m10^{-4} m B) Cu(aq)2+∣Cu(s)Cu^{2+}_{(aq)} | Cu_{(s)} 10−1M10^{-1}M C) Cu(aq)2+∣Cu(s)Cu^{2+}_{(aq)} | Cu_{(s)} 10−2M10^{-2}M

If the standard reduction potential of Cu is 0.34 V, reduction potential in volts of the above follows the order

A

A > C > B

B

B > C > A

C

A > B > C

D

C > B > A

Answer

B > C > A

Explanation

Solution

The reduction potential of an electrode is given by the Nernst equation: Ered=Ered0−0.0591nlog⁡1[Cu2+]E_{red} = E^0_{red} - \frac{0.0591}{n} \log \frac{1}{[Cu^{2+}]}. Given Ered0=0.34E^0_{red} = 0.34 V and n=2n=2. For electrode A: [Cu2+]=10−4[Cu^{2+}] = 10^{-4} M. EA=0.34−0.05912log⁡110−4=0.34−0.05912×4=0.34−0.1182=0.2218E_A = 0.34 - \frac{0.0591}{2} \log \frac{1}{10^{-4}} = 0.34 - \frac{0.0591}{2} \times 4 = 0.34 - 0.1182 = 0.2218 V. For electrode B: [Cu2+]=10−1[Cu^{2+}] = 10^{-1} M. EB=0.34−0.05912log⁡110−1=0.34−0.05912×1=0.34−0.02955=0.31045E_B = 0.34 - \frac{0.0591}{2} \log \frac{1}{10^{-1}} = 0.34 - \frac{0.0591}{2} \times 1 = 0.34 - 0.02955 = 0.31045 V. For electrode C: [Cu2+]=10−2[Cu^{2+}] = 10^{-2} M. EC=0.34−0.05912log⁡110−2=0.34−0.05912×2=0.34−0.0591=0.2809E_C = 0.34 - \frac{0.0591}{2} \log \frac{1}{10^{-2}} = 0.34 - \frac{0.0591}{2} \times 2 = 0.34 - 0.0591 = 0.2809 V. Comparing the potentials: EB(0.31045 V)>EC(0.2809 V)>EA(0.2218 V)E_B (0.31045 \text{ V}) > E_C (0.2809 \text{ V}) > E_A (0.2218 \text{ V}). Therefore, the order is B > C > A.