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)EAg+/Ag°E^{\degree}_{Ag^{+/Ag}}=+ 0.80 V
Then reduction potential in volts of the electrodes in the order

Updated On: Mar 29, 2025
  • B > C > D > A
  • C > D > A > B
  • A > D > C > B
  • A > B > C > D
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The Correct Option is A

Solution and Explanation

The given electrodes and their standard reduction potentials are:

  • A: Ag+(0.0001M)/Ag(s) \text{Ag}^+ (0.0001\text{M})/\text{Ag}(s)
  • B: Ag+(0.1M)/Ag(s) \text{Ag}^+ (0.1\text{M})/\text{Ag}(s)
  • C: Ag+(0.01M)/Ag(s) \text{Ag}^+ (0.01\text{M})/\text{Ag}(s)
  • D: Ag+(0.001M)/Ag(s) \text{Ag}^+ (0.001\text{M})/\text{Ag}(s) , EAg+/Ag=+0.80 V 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=E0.0591nlog([Reductant][Oxidant]) 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 n = 1 ), the equation simplifies to: E=E0.0591log[Ag+] E = E^\circ - 0.0591 \log [\text{Ag}^+] The standard reduction potential EAg+/Ag 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: EA=0.800.0591log(0.0001)=0.800.0591×(4)=0.80+0.2364=1.0364 V 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: EB=0.800.0591log(0.1)=0.800.0591×(1)=0.80+0.0591=0.8591 V 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: EC=0.800.0591log(0.01)=0.800.0591×(2)=0.80+0.1182=0.9182 V 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: ED=0.800.0591log(0.001)=0.800.0591×(3)=0.80+0.1773=0.9773 V 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: B>C>D>A \text{B} > \text{C} > \text{D} > \text{A} Therefore, the correct order is Option (A): B > C > D > A.

Correct Answer: Option (A)

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