Consider the cell:Pt_{(s)} | H_{2(g)}(1 atm) | vedclass

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

Question

(B) The cell reaction is: $\frac{1}{2}H_{2(g)} + Fe^{3+}_{(aq)} \longrightarrow H^{+}_{(aq)} + Fe^{2+}_{(aq)}$The Nernst equation for the cell is: $E_

Vedclass · Accepted Answer

$10$

Vedclass · Answer

(B) The cell reaction is: $\frac{1}{2}H_{2(g)} + Fe^{3+}_{(aq)} \longrightarrow H^{+}_{(aq)} + Fe^{2+}_{(aq)}$The Nernst equation for the cell is: $E_{cell} = E^0_{cell} - \frac{0.0591}{n} \log \frac{[H^{+}][Fe^{2+}]}{[Fe^{3+}](P_{H_2})^{1/2}}$Here,$n = 1$,$[H^{+}] = 1 \ M$,$P_{H_2} = 1 \ atm$,and $E^0_{cell} = E^0_{cathode} - E^0_{anode} = 0.771 \ V - 0 \ V = 0.771 \ V$.Substituting the values: $0.712 = 0.771 - 0.0591 \log \frac{[Fe^{2+}]}{[Fe^{3+}]}$$0.0591 \log \frac{[Fe^{2+}]}{[Fe^{3+}]} = 0.771 - 0.712 = 0.059$$\log \frac{[Fe^{2+}]}{[Fe^{3+}]} = \frac{0.059}{0.0591} \approx 1$Therefore,$\frac{[Fe^{2+}]}{[Fe^{3+}]} = 10^1 = 10$.

Similar Questions

For a general redox reaction: $aA + bB \xrightarrow{n e^-} cC + dD$. Derive the Nernst equation.

Calculate the e.m.f. of the half-cell given below: $Fe | FeSO_4$ $(a = 0.1 \ M)$ where $E^o_{OP} = 0.44 \ V$. (in $V$)

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

What would be the electrode potential for the given half-cell reaction at $pH = 5$ (in $V$)?
$2 H_2O \rightarrow O_2 + 4 H^{\oplus} + 4 e^{-} ; E_{red}^{0} = 1.23 \ V$
$(R = 8.314 \ J \ mol^{-1} \ K^{-1} ; \text{Temperature} = 298 \ K ; \text{oxygen under std. atm. pressure of } 1 \ bar)$

Vedclass Test Series

Exam Paper Generator

Online Exam Module