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)