Which of the following expressions is used to calculate $E_{cell}$ for the following cell at $25^{\circ} C$?
$Pb_{(s)} | Pb^{2+}_{(1 \ M)} || Ag^{+}_{(10 \ M)} | Ag_{(s)}$
$Pb_{(s)} | Pb^{2+}_{(1 \ M)} || Ag^{+}_{(10 \ M)} | Ag_{(s)}$
- A$E_{cell} = (E^{\circ}_{cell} + 0.0592) \ V$
- B$E_{cell} = (E^{\circ}_{cell} - 0.0592) \ V$
- C$E_{cell} = (E^{\circ}_{cell} - 0.0296) \ V$
- D$E_{cell} = (E^{\circ}_{cell} + 0.0296) \ V$
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Similar Questions
$Pt_{(s)} | H_{2(g)}(1 \ bar) | H^{+}_{(aq)}(1 \ M) || M^{3+}_{(aq)}, M^{+}_{(aq)} | Pt_{(s)}$
The $E_{cell}$ for the given cell is $0.1115 \ V$ at $298 \ K$ when $\frac{[M^{+}_{(aq)}]}{[M^{3+}_{(aq)}]} = 10^{a}$.
The value of $a$ is.
Given : $E^{\circ}_{M^{3+}/M^{+}} = 0.2 \ V$
$\frac{2.303 \ RT}{F} = 0.059 \ V$
The $E_{cell}$ for the given cell is $0.1115 \ V$ at $298 \ K$ when $\frac{[M^{+}_{(aq)}]}{[M^{3+}_{(aq)}]} = 10^{a}$.
The value of $a$ is.
Given : $E^{\circ}_{M^{3+}/M^{+}} = 0.2 \ V$
$\frac{2.303 \ RT}{F} = 0.059 \ V$
DifficultJEE MAIN 2023
View Solution$A$ hydrogen electrode is made by dipping platinum wire in a solution of nitric acid of $pH=9$ and passing hydrogen gas around the platinum wire at $1.2 \ atm$ pressure. The oxidation potential of such an electrode equals $V$.
Which of the following is correct as a Nernst equation for the given electrochemical cell ?
$Mg_{(s)}|Mg_{(aq)}^{2+}(0.1 \ M)||Cl_{(aq)}^{-}(0.1 \ M)|Cl_{2_{(g)}}(1 \ bar)|Pt_{(s)}$
$Mg_{(s)}|Mg_{(aq)}^{2+}(0.1 \ M)||Cl_{(aq)}^{-}(0.1 \ M)|Cl_{2_{(g)}}(1 \ bar)|Pt_{(s)}$
The correct representation of Nernst's equation for the reduction of a metal ion $M^{n+}$ to metal $M$ is:
Easy
View SolutionWhat is the change in the reduction potential of a hydrogen electrode when the $pH$ of the initial solution changes from $0$ to $7$ by neutralization?
Medium
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