Aluminum displaces hydrogen in dil. HCl whereas silver does not. The e.m.f of a cell prepared by combination $A l / {A l}^{3 +} a n d A g / {A g}^{+}$ is 2.46 V. The reduction potential reduction potential of aluminum electrodes is:

+1.66 V

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-3.26 V

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+3.26 V

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-1.66 V

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Solution

The correct option is A -1.66 V
$A l + 3 {A g}^{+} \to 3 A g + {A l}^{3 +}$
i.e. $E_{c e l l} = E_{{A g}^{+} / A g} - E_{{A l}^{3 +} / A l}$
we know, $E_{{A g}^{+} / A g} = 0.80 V$
$E_{c e l l} = 2.46 V$
$2.46 = 0.80 V - E_{{A l}^{3 +} / A l}$
$∴ E_{{A l}^{3 +} / A l} = - 1.66 V$

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Similar questions

Q. Aluminum displace hydrogen from dilute

H C l

whereas silver does not. The emf of a cell prepared by combining

A l / A l^{3 +}

and

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2.46 V

. The reduction potential of the silver electrode is

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. The reduction potential of aluminum electrode is:

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Aluminum displaces hydrogen in dil. HCl whereas silver does not. The e.m.f of a cell prepared by combination Al/Al3+ and Ag/Ag+ is 2.46 V. The reduction potential reduction potential of aluminum electrodes is:

The correct option is A -1.66 VAl+3Ag+→3Ag+Al3+i.e. Ecell=EAg+/Ag−EAl3+/Alwe know, EAg+/Ag=0.80VEcell=2.46V2.46=0.80V−EAl3+/Al∴EAl3+/Al=−1.66V

Aluminum displaces hydrogen in dil. HCl whereas silver does not. The e.m.f of a cell prepared by combination $A l / {A l}^{3 +} a n d A g / {A g}^{+}$ is 2.46 V. The reduction potential reduction potential of aluminum electrodes is:

The correct option is A -1.66 V
$A l + 3 {A g}^{+} \to 3 A g + {A l}^{3 +}$
i.e. $E_{c e l l} = E_{{A g}^{+} / A g} - E_{{A l}^{3 +} / A l}$
we know, $E_{{A g}^{+} / A g} = 0.80 V$
$E_{c e l l} = 2.46 V$
$2.46 = 0.80 V - E_{{A l}^{3 +} / A l}$
$∴ E_{{A l}^{3 +} / A l} = - 1.66 V$