For the following reactions-Zn rightarrow Zn-2- - 2e- E-0-76 VAu rightarrow Au-3- - 3e- E-1-42 VIf gold foil is placed in a solution containing Zn-2- the reaction potential would be-1-34 V-2-18 V-0-66 V-1-34 V-2-18 V

Question

Toppr Admin · Accepted Answer

If gold foil is placed in a solution containing Zn2- Gold will be reduced and Zinc will be oxidised and the following reaction will take place-3Zn-2Au3-x2192-3Zn2-2AuThe reaction potential for the above reaction is given by -E-Ecathode-x2212-Eanode-x2212-1-42-x2212-0-76-x2212-2-18-xA0-V

For the following reactions:Zn→Zn2++2e−;E=+0.76VAu→Au3++3e−;E=−1.42VIf gold foil is placed in a solution containing Zn2+, the reaction potential would be:−1.34V−2.18V−0.66V+1.34V+2.18V

If gold foil is placed in a solution containing Zn2+, Gold will be reduced and Zinc will be oxidised and the following reaction will take place:3Zn+2Au3+→3Zn2++2AuThe reaction potential for the above reaction is given by :E=Ecathode−Eanode=−1.42−0.76=−2.18 V

For the following reactions:
$Z n \to Z n^{2 +} + 2 e^{-}; E = + 0.76 V$
$A u \to A u^{3 +} + 3 e^{-}; E = - 1.42 V$
If gold foil is placed in a solution containing $Z n^{2 +}$ , the reaction potential would be:
$- 1.34 V$
$- 2.18 V$
$- 0.66 V$
$+ 1.34 V$
$+ 2.18 V$