Pt_{(s)} | H_{2(g)}(1 bar) | H^{+}_{(aq)}(1 | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The cell reaction is:$H_{2(g)} + M^{3+}_{(aq)} \longrightarrow 2H^{+}_{(aq)} + M^{+}_{(aq)}$Using the Nernst equation:$E_{cell} = E^{\circ}_{cell}

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(D) The cell reaction is:$H_{2(g)} + M^{3+}_{(aq)} \longrightarrow 2H^{+}_{(aq)} + M^{+}_{(aq)}$Using the Nernst equation:$E_{cell} = E^{\circ}_{cell} - \frac{0.059}{n} \log Q$Here,$n = 2$ (number of electrons transferred).$E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode} = 0.2 \ V - 0 \ V = 0.2 \ V$.Substituting the values:$0.1115 = 0.2 - \frac{0.059}{2} \log \frac{[M^{+}] [H^{+}]^2}{[M^{3+}] P_{H_2}}$Since $[H^{+}] = 1 \ M$ and $P_{H_2} = 1 \ bar$:$0.1115 = 0.2 - 0.0295 \log \frac{[M^{+}]}{[M^{3+}]}$$0.0885 = 0.0295 \log \frac{[M^{+}]}{[M^{3+}]}$$\log \frac{[M^{+}]}{[M^{3+}]} = \frac{0.0885}{0.0295} = 3$Given $\frac{[M^{+}]}{[M^{3+}]} = 10^{a}$,so $10^a = 10^3$,which implies $a = 3$.

Similar Questions

Write the Nernst equation for the $E_{cell}$ of a Daniell cell.

In which of the following conditions will the reduction potential of a hydrogen half-cell be negative?

At $298 \ K$ the equilibrium constant for the reaction $M_{(s)} + 2 Ag^{+}_{(aq)} \rightarrow M^{2+}_{(aq)} + 2 Ag_{(s)}$ is $10^{15}$. What is the $E_{cell}^{\ominus}$ (in $V$) for this reaction? $\left(\frac{2.303 RT}{F}\right) = 0.06 \ V$

$Cu_{(s)} | Cu^{+2}(aq, 10^{-3} M) || Ag^{+}(aq, 10^{-5} M) | Ag_{(s)}$
If $E^{o}_{Cu^{+2}/Cu} = +0.34 \ V$
$E^{o}_{Ag^{+}/Ag} = +0.80 \ V$
$E_{cell}$ will be

Consider a Daniell cell operating under non-standard state conditions. Suppose that the cell's reaction is multiplied by $2$. Which of the following will double its initial value?

Vedclass Test Series

Exam Paper Generator

Online Exam Module