Consider the half-reaction \(\mathrm{Ag}^{+}(a q)+\mathrm{e}^{-} \longrightarrow \mathrm{Ag}(s)\). Which of the lines in the following diagram indicates how the reduction potential varies with the concentration of \(\mathrm{Ag}^{+} ?(\mathrm{~b})\) What is the value of \(E_{\mathrm{red}}\) when \(\log \left[\mathrm{Ag}^{+}\right]=0 ?\) [Section \(\left.20.6\right]\)

The Nernst Equation provides the relationship between the standard reduction potential (E°) and the actual reduction potential (E) for a half-reaction under non-standard conditions. This equation is: \[E = E° - \frac{RT}{nF} \ln Q\] Where: - E° is the standard reduction potential - R is the gas constant (8.314 J/(mol·K)) - T is the temperature of the reaction (in Kelvin) - n is the number of moles of electrons exchanged in the half-reaction - F is Faraday's constant (96485 C/mol) - Q is the reaction quotient (concentration of products to reactants) For our case, the half-reaction is given by: \[\mathrm{Ag}^{+}(a q)+\mathrm{e}^{-} \longrightarrow \mathrm{Ag}(s)\] Here, n = 1 (i.e., only one electron is exchanged in the half-reaction), and the reaction quotient, Q, is simply equal to the concentration of Ag+ ions, [Ag+].

We can rewrite the Nernst Equation for our specific half-reaction as: \[E = E° - \frac{RT}{F} \ln [\mathrm{Ag}^{+}]\] As we are concerned with log[Ag+], we can update the equation by applying the property of logarithm as follows: \[E = E° - \frac{2.303 RT}{F} \log [\mathrm{Ag}^{+}]\] Now, we are given \(\log [\mathrm{Ag}^{+}]= 0\), which means that the concentration of Ag+ [Ag+] = 1 (as the base 10 logarithm of 1 equals 0).

Substitute the given log[Ag+] value into the equation: \[E = E° - \frac{2.303 RT}{F} \cdot 0\]

Since anything multiplied by 0 equals 0, the equation becomes: \[E = E°\] In this case, when \(\log [\mathrm{Ag}^{+}] = 0\), the actual reduction potential (E_red;under non-standard condition) is equal to the standard reduction potential (E°) for the given half-reaction.