For the cell reaction,$Zn_{(s)} + 2 Ag_{(aq)}^{+} \longrightarrow Zn_{(aq)}^{2+} + 2 Ag_{(s)}$,the cell potential is less than $E^{\circ}_{cell}$ by $0.0592 \ V$ at $298 \ K$ when:
- A$[Zn^{2+}] = 1 \ M$ and $[Ag^{+}] = 0.1 \ M$
- B$[Zn^{2+}] = 1 \ M$ and $[Ag^{+}] = 0.01 \ M$
- C$[Zn^{2+}] = 0.1 \ M$ and $[Ag^{+}] = 1 \ M$
- D$[Zn^{2+}] = 0.01 \ M$ and $[Ag^{+}] = 1 \ M$
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$Cu_{(s)} | Cu^{+2}(aq, 10^{-3} M) || Ag^{+}(aq, 10^{-5} M) | Ag_{(s)}$
If $E^{o}_{Cu^{+2}/Cu} = +0.34 \ V$
$E^{o}_{Ag^{+}/Ag} = +0.80 \ V$
$E_{cell}$ will be
If $E^{o}_{Cu^{+2}/Cu} = +0.34 \ V$
$E^{o}_{Ag^{+}/Ag} = +0.80 \ V$
$E_{cell}$ will be
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View SolutionThe standard emf for the cell $Cd_{(s)}|Cd^{2+}_{(aq)}(1 \ M)||Cu^{2+}_{(aq)}(1 \ M)|Cu_{(s)}$ is $0.74 \ V$. If the concentration of $Cd^{2+}_{(aq)}$ and $Cu^{2+}_{(aq)}$ both decrease by $10$ times at $298 \ K$,calculate the emf of the cell.
The Gibbs energy change of the reaction (in $kJ \ mol^{-1}$) corresponding to the following cell $Cr | Cr^{3+} (0.1 \ M) || Fe^{2+} (0.01 \ M) | Fe$ is: (Given: $E^{\circ}_{Cr^{3+}/Cr} = -0.74 \ V$,$E^{\circ}_{Fe^{2+}/Fe} = -0.44 \ V$)
What pressure $(bar)$ of $H_2$ would be required to make the $emf$ of a hydrogen electrode zero in pure water at $25^{\circ} C$?
The $EMF$ of a hydrogen electrode in terms of $pH$ is (at $1 \ atm$ pressure).
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