$Pt_{(s)} | H_{2(g)} (1 \ atm) | H^{+} (pH = 2) || H^{+} (pH = 3) | H_{2(g)} (1 \ atm) | Pt_{(s)}$ cell reaction will be

  • A
    spontaneous
  • B
    nonspontaneous
  • C
    equilibrium
  • D
    None of these

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Similar Questions

At $298 \ K$,some standard electrode potentials are given below:
$Pb^{2+} / Pb$$-0.13 \ V$
$Ni^{2+} / Ni$$-0.24 \ V$
$Cd^{2+} / Cd$$-0.40 \ V$
$Fe^{2+} / Fe$$-0.44 \ V$

Metal rods $X$ and $Y$ are inserted into a solution containing $0.001 \ M$ $X^{2+}$ and $0.1 \ M$ $Y^{2+}$ at $298 \ K$ and connected by a conducting wire. This results in the dissolution of $X$. The correct combination$(s)$ of $X$ and $Y$ are,respectively:
(Given: Gas constant,$R = 8.314 \ J \ K^{-1} \ mol^{-1}$,Faraday constant,$F = 96500 \ C \ mol^{-1}$)
$(A) \ Cd$ and $Ni \ \ (B) \ Cd$ and $Fe \ \ (C) \ Ni$ and $Pb \ \ (D) \ Ni$ and $Fe$

The standard $EMF$ for the given cell reaction $Zn + Cu^{2+} \rightarrow Cu + Zn^{2+}$ is $1.10 \ V$ at $25^oC$. The $EMF$ for the cell reaction,when $0.1 \ M \ Cu^{2+}$ and $0.1 \ M \ Zn^{2+}$ solutions are used,at $25^oC$ is .......... $V$.

Calculate the cell potential for the cell $Zn_{(s)} | Zn^{2+} (0.6 \ M) || Cd^{2+} (0.85 \ M) | Cd_{(s)}$ at $298 \ K$. (Given: $E^{\circ}_{Zn^{2+}/Zn} = -0.76 \ V$ and $E^{\circ}_{Cd^{2+}/Cd} = -0.40 \ V$) (in $V$)

Calculate $\Delta G$ and $E_{cell}$ for the following cell at $298 \ K$ temperature.
$Al_{(s)} | Al^{3+} (0.01 \ M) || Fe^{2+} (0.02 \ M) | Fe_{(s)}$ $\left[ E^o_{Al^{3+}|Al} = -1.66 \ V \right.$ and $\left. E^o_{Fe^{2+}|Fe} = -0.44 \ V \right]$

The standard $emf$ for the cell reaction $Zn_{(s)} + Cu^{2+}_{(aq)} \rightarrow Zn^{2+}_{(aq)} + Cu_{(s)}$ is $1.10 \, V$ at $25 \, ^\circ C$. What will be the $emf$ of the cell when using $0.1 \, M \, Cu^{2+}$ and $0.1 \, M \, Zn^{2+}$ solutions (in $, V$)?

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