Calculate the standard cell potential for a cell having the following reaction: $2 Al_{(s)} + 3 Ni^{2+} \rightarrow 2 Al^{3+} + 3 Ni_{(s)}$ given that $E_{Ni^{2+}/Ni}^{\circ} = -0.25 \ V$ and $E_{Al^{3+}/Al}^{\circ} = -1.66 \ V$. (in $V$)
- A$0.50$
- B$1.41$
- C$-0.50$
- D$0.41$
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The standard electrode potential $(E^{\circ})$ values of $Al^{3+}/Al, Ag^{+}/Ag, K^{+}/K$ and $Cr^{3+}/Cr$ are $-1.66 \ V, 0.80 \ V, -2.93 \ V$ and $-0.74 \ V,$ respectively. The correct decreasing order of reducing power of the metal is:
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View Solution$A$ solution contains $Fe^{2+}$,$Fe^{3+}$ and $I^{-}$ ions. This solution was treated with iodine at $35^{\circ}C$. $E^{\circ}$ for $Fe^{3+}/Fe^{2+}$ is $+0.77 \ V$ and $E^{\circ}$ for $I_2/2I^{-}$ is $+0.536 \ V$. The favourable redox reaction is
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View SolutionWhen a rod of metal $A$ is dipped in an aqueous solution of metal $B$ (concentration of $B^{2+}$ ion being $1 \ M$) at $25 \ ^oC$,the standard electrode potentials are $E^o_{A^{2+}/A} = -0.76 \ V$ and $E^o_{B^{2+}/B} = +0.34 \ V$. What will happen?
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View SolutionWhich metal cannot produce $H_2$ gas by reaction with $HCl$ solution?
$E^0_{Fe^{2+}/Fe} = -0.44 \ V$
$E^0_{Cu^{2+}/Cu} = +0.34 \ V$
$E^0_{Ni^{2+}/Ni} = -0.25 \ V$
$E^0_{Zn^{2+}/Zn} = -0.76 \ V$
$E^0_{Fe^{2+}/Fe} = -0.44 \ V$
$E^0_{Cu^{2+}/Cu} = +0.34 \ V$
$E^0_{Ni^{2+}/Ni} = -0.25 \ V$
$E^0_{Zn^{2+}/Zn} = -0.76 \ V$
Given at $298 \ K$: $E^\ominus_{Fe^{2+}/Fe} = X \ V$; $E^\ominus_{Fe^{3+}/Fe} = Y \ V$. The $E^\ominus_{Fe^{3+}/Fe^{2+}}$ in Volt at $298 \ K$ is given by:
DifficultJEE MAIN 2026
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