Question

Tollen's reagent is used for the detection of aldehyde when a solution of $AgN{O}_3$ is added to glucose with $N{H}_{4}OH$ then gluconic acid is formed
$A{g}^{+}+e^{-}\to Ag;E_{red}^{+}=0.8V$
$C_6{H}_{12}{O}_6+H_{2}O\to$ Gluconic acid $\left(C_6{H}_{12}{O}_7\right)+2H^{+}+2e^{-}$
$E_{red}{}^{\circ }=-0.05V$
$Ag{\left(N{H}_3\right)}_2{}^{+}+e^{-}\to Ag(s)+2N{H}_3;E_{red}{}^{\circ }=0.337V$
[Use $2.303\times \frac{RT}{F}=0.0592$ and $\frac{F}{RT}=38.92$ at $398K]$
When ammonia is added to the solution, $pH$ is raised to $11$ . Which half- cell reaction is affected by $pH$ and by how much?

• A. $E_{\text{oxd }}$ will increase by a factor of $0.65$ from $E_{\text{oxd }}{}^{+}$
• B. $E_{\text{oxd }}$ will decrease by a factor of $0.65$ from $E_{\text{oxd }}^{+}$
• C. $E_{\text{red }}$ will increase by a factor of $0.65$ from $E_{\text{red }}{}^{+}$
• D. $E_{\text{red }}$ will decrease by a factor of $0.65$ from $E_{\text{red }}^{+}$

Answer 1

Answer: (C)

The expression for oxidation reaction is
$E_{\text{oxd }}=E_{\text{oxd }}{}^{\circ }-\frac{2.303RT}{nF}\log \frac{\left[C_6{H}_{12}{O}_6\right]}{\left[C_6{H}_{12}{O}_7\right]{\left[H^{+}\right]}^2}$
The expression for reduction is
$E_{red}=E_{red}{}^{\circ }-\frac{2.303RT}{nF}\log \frac{1}{\left[A{g}^{+}\right]}$
The addition of $N{H}_3$ results in formation of more ${\left[Ag{\left(N{H}_3\right)}_2\right]}^{+}$ complex, thus reducing the concentration of $A{g}^{+}$ions available for reduction. Thus addition of $N{H}_3$ causes concentration of $\left[H^{+}\right]$to decrease and $pH$ to increase. So $E_{red}$ increases approximately by a factor of $0.65V$.
$E_{red}=E_{red}{}^{\circ }-\frac{2.303RT}{nF}\log {\left[H^{+}\right]}^2$
$=0-\frac{0.0592}{2}\times 2\times 11=0.6512$