The following data (in kJ mol^-1) are given: for Zn, Delta_aH = 130, Delta_iH_1 + Delta_iH_2 = 2640, Delta_hydH(Zn^2+) = -2046; for Cu, Delta_aH = 338, Delta_iH_1 + Delta_iH_2 = 2704, Delta_hydH(Cu^2+) = -2121. The enthalpy change for M(s) -> M^2+(aq) + 2e^- is therefore

Cu: +921, Zn: +724; Cu is the weaker reducing agent

Cu: +921, Zn: +724; Zn is the weaker reducing agent

Cu: +5163, Zn: +4816; Zn is the weaker reducing agent

Cu: +921, Zn: +724; Cu is the weaker reducing agent

Cu: +583, Zn: +594; Cu is the stronger reducing agent

JEE Advanced levelSingle correct optionDifficulty 3 of 3df073

The following data (in $kJ mo l^{- 1}$ ) are given: for Zn, $Δ_{a} H = 130$ , $Δ_{i} H_{1} + Δ_{i} H_{2} = 2640$ , $Δ_{hyd} H (Zn X^{2 +}) = - 2046$ ; for Cu, $Δ_{a} H = 338$ , $Δ_{i} H_{1} + Δ_{i} H_{2} = 2704$ , $Δ_{hyd} H (Cu X^{2 +}) = - 2121$ . The enthalpy change for $M (s) \to M X^{2 +} (aq) + 2 e^{-}$ is therefore

The enthalpy change is

Δ_{a} H + Δ_{i} H + Δ_{hyd} H

(hydration is exothermic). For Zn:

130 + 2640 - 2046 = + 724

. For Cu:

338 + 2704 - 2121 = + 921

. Oxidising Cu costs 197 kJ/mol more, which is consistent with Cu having the positive

E^{\circ}

and Zn the negative one, so Cu is the weaker reductant.