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

Calculate the cell potential for a $Cu$ plate kept in $0.2 \ M$ $CuSO_4$ solution. Given: $E_{Cu^{2+} \mid Cu}^o = 0.34 \ V$. (in $V$)

The correct representation of Nernst's equation for the reduction of a metal ion $M^{n+}$ to metal $M$ is:

For the galvanic cell,
$Zn_{(s)} + Cu^{2+}(0.02 \ M) \rightarrow Zn^{2+}(0.04 \ M) + Cu_{(s)}$
$E_{cell} = ...... \times 10^{-2} \ V \text{ (Nearest integer) }$
$[\text{Use}: E_{Cu^{2+}/Cu}^{0} = 0.34 \ V, E_{Zn^{2+}/Zn}^{0} = -0.76 \ V]$
$[\frac{2.303 \ RT}{F} = 0.059 \ V]$

Calculate the cell potential at $298 \ K$ for the following cell:
$Cu_{(s)} | Cu^{2+}(0.1 \ M) || Cu^{2+}(1 \ M) | Cu_{(s)}$
Given: $E_{Cu^{2+}|Cu}^o = 0.34 \ V$ (in $V$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module