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.

$A{g}^{+}+e^{-}\to Ag;$ $E_{red}^{\circ }=0.8V$

$C_6{H}_{12}{O}_6+H_{2}O\to C_6{H}_{12}{O}_7$ (Gluconic acid) $+2H^{+}+2e;E_{oxd}^{\circ }=-0.05V$

$\left[Ag\right(N{H}_3{)}_2{]}^{+}+e^{-}\to Ag\left(S\right)+2N{H}_3;E_{red}^{\circ }=–0.337V$

[Use $2.303\times \frac{RT}{F}=0.0592$ and $\frac{F}{RT}=38.92$ at 298 K]

(ii) 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. $E_{oxd}$ will increase by 0.65 from $E_{oxd}^{\circ }$
• B. $E_{oxd}$ will decrease by 0.65 from $E_{oxd}^{\circ }$
• C. $E_{red}$ will increase by 0.65 from $E_{red}^{\circ }$
• D. $E_{red}$ will decrease by 0.65 from $E_{red}^{\circ }$

Answer 1

The correct option is C

$E_{red}$ will increase by 0.65 from $E_{red}^{\circ }$

On increasing the concentration $N{H}_3$ the concentration of $H^{+}$ ion decreases. Hence:

$E_{red}=E_{red}^{\circ }-\frac{0.0592}{2}log\left(\right[H^{+}{]}^2)$

$=0-\frac{0.0592}{2}\times 2log\left({10}^{-11}\right)$ = 0.65V

Thus, $E_{red}$ increases by 0.65V