Aluminium displaces hydrogen from acids, but copper does not. A galvanic cell prepared by combining $C u / C u^{2 +}$ and $A I / A I^{3 +}$ has an emf of 2.0 V at 298 K. If the potential of copper electrode is + 0.34 V, that of Aluminium electrons is?

-2.3 V

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+2.34 V

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-1.66 V

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1.66 V

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Solution

The correct option is B -1.66 V
at anode,
$3 C u \to 3 C u^{+ 2} + 6 e^{-}$
at cathode,
$2 A l^{+ 3} + 6 r^{-} \to 2 A l_{(3)}$

$3 C u_{(s)} + 2 A l^{+ 3} \to 3 C u^{+ 2} + 2 A l_{(s)}$
$E_{c e l l} = 2.0 V$
$E_{C u^{+ 2} / C u} = 0.34 V$

$E_{c e l l} = E_{C u^{+ 2} / C u} - E_{A l^{+ 3} / A l}$

$2 = 0.34 - E_{A l^{+ 3} / A l}$

$E_{A l^{+ 3} / A l} = 0.34 - 2$

$E_{A l^{+ 3} / A l} = - 1.66$
therefore the correct Answer is option $C$ $- 1.66$

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Aluminium displaces hydrogen from acids, but copper does not. A galvanic cell prepared by combining Cu/Cu2+ and AI/AI3+ has an emf of 2.0 V at 298 K. If the potential of copper electrode is + 0.34 V, that of Aluminium electrons is?

The correct option is B -1.66 Vat anode,3Cu→3Cu+2+6e−at cathode,2Al+3+6r−→2Al(3)3Cu(s)+2Al+3→3Cu+2+2Al(s)Ecell=2.0VECu+2/Cu=0.34VEcell=ECu+2/Cu−EAl+3/Al2=0.34−EAl+3/AlEAl+3/Al=0.34−2EAl+3/Al=−1.66therefore the correct Answer is option C −1.66

Aluminium displaces hydrogen from acids, but copper does not. A galvanic cell prepared by combining $C u / C u^{2 +}$ and $A I / A I^{3 +}$ has an emf of 2.0 V at 298 K. If the potential of copper electrode is + 0.34 V, that of Aluminium electrons is?