Question

Which of the following represents the potential of silver wire dipper into 0.1 M AgNO$_3$ solution at 25 $^\circ$C?
A) E$^\circ$$_{red}$
B) (E$^\circ$$_{red}$ +0.059)
C) (E$^\circ$$_{oxid}$ -0.059)
D) (E$^\circ$$_{red}$ -0.059)

Answer 1

Hint: The cell potential is calculated with the help of Nernst equation. In this question, we will use the Nernst equation for the oxidation, and reduction part of silver. Then, the correct representation can be determined.

Complete step by step answer:
*First, let us write the Nernst equation at 25$^\circ$C, or 298 K; and the equation represents the dependence of total potential on the reaction quotient.
*The representation is
E$_{cell}$ = E$^\circ$$_{cell}$ $\dfrac{0.0591}{n}$ log $\dfrac{[Ag]}{[Ag^{+}]}$
 *In this representation, log $\dfrac{[Ag]}{[Ag^{+}]}$ represents log Q; i.e. reaction quotient, numerator in the fraction represents the oxidation, and denominator represents the reduction.
*If we talk about the other terms, E$_{cell}$ is cell potential of the cell, and E$^\circ$$_{cell}$ is considered to be the cell potential under standard conditions.
*Now, n represents the charge, so in this equation n is 1 for silver wire, as silver has +1 positive charge.
*The chemical equation will be
Ag $^{+}$ + e$^{-}$ $\rightarrow$ Ag
*Thus, E$_{cell}$ = E$^\circ$$_{cell}$ $\dfrac{0.0591}{1}$ log $\dfrac{10}{1}$; here we have considered the value given for silver nitrate solution i.e. 0.1 M.
*Then, we have, E$_{cell}$ = E$^\circ$$_{red}$ -0.0591 log 10,
E$_{cell}$ = E$^\circ$$_{red}$ -0.0591
*So, the potential of silver wire dipper into 0.1 M silver nitrate solution at 25$^\circ$C is represented by E$^\circ$$_{red}$ -0.059.
Hence, the correct option is (D).

Note: We considered here the cell potential for reduction while representing the overall potential of the cell as mentioned. As in the chemical equation we can see that the silver is gaining an electron; it means that silver acts as an oxidizing agent, and the reduction process takes place. Thus, there is no need of confusion.