Question

Consider the following 4 electrodes
A : Ag⁺ (0.0001 M)/Ag_(s) ;  B : Ag⁺ (0.1 M) / Ag_(s)
C : Ag⁺ (0.01 M)/Ag_(s) ;        D : Ag⁺ (0.001 M) / Ag_(s); \(E^{\degree}_{Ag^{+/Ag}}\)=+ 0.80 V
Then reduction potential in volts of the electrodes in the order

• A. B > C > D > A
• B. C > D > A > B
• C. A > D > C > B
• D. A > B > C > D

Answer 1

The Correct Option is A

The given electrodes and their standard reduction potentials are:

• A: \( \text{Ag}^+ (0.0001\text{M})/\text{Ag}(s) \)
• B: \( \text{Ag}^+ (0.1\text{M})/\text{Ag}(s) \)
• C: \( \text{Ag}^+ (0.01\text{M})/\text{Ag}(s) \)
• D: \( \text{Ag}^+ (0.001\text{M})/\text{Ag}(s) \), \( E^\circ_{\text{Ag}^+/\text{Ag}} = +0.80 \text{ V} \)

Explanation:

The reduction potential for the given electrodes can be calculated using the Nernst equation: \[ E = E^\circ - \frac{0.0591}{n} \log \left( \frac{[\text{Reductant}]}{[\text{Oxidant}]} \right) \] Since the number of electrons involved in the silver (Ag) reduction reaction is 1 (i.e., \( n = 1 \)), the equation simplifies to: \[ E = E^\circ - 0.0591 \log [\text{Ag}^+] \] The standard reduction potential \( E^\circ_{\text{Ag}^+/\text{Ag}} \) is given as +0.80 V for standard conditions.

Step 1: Calculating Reduction Potentials:

Using the Nernst equation: - For electrode A: \[ E_A = 0.80 - 0.0591 \log (0.0001) = 0.80 - 0.0591 \times (-4) = 0.80 + 0.2364 = 1.0364 \text{ V} \] - For electrode B: \[ E_B = 0.80 - 0.0591 \log (0.1) = 0.80 - 0.0591 \times (-1) = 0.80 + 0.0591 = 0.8591 \text{ V} \] - For electrode C: \[ E_C = 0.80 - 0.0591 \log (0.01) = 0.80 - 0.0591 \times (-2) = 0.80 + 0.1182 = 0.9182 \text{ V} \] - For electrode D: \[ E_D = 0.80 - 0.0591 \log (0.001) = 0.80 - 0.0591 \times (-3) = 0.80 + 0.1773 = 0.9773 \text{ V} \]

Conclusion:

The reduction potentials for the electrodes in descending order are: \[ \text{B} > \text{C} > \text{D} > \text{A} \] Therefore, the correct order is Option (A): B > C > D > A.

Correct Answer: Option (A)