Use appropriate data to calculate δg∘ for the reaction. 2mno−4(aq)+cd(s)→2mno2−4(aq)+cd2+(a q) express your answer to three significant figures and include the appropriate units.

Reaction at anode: Cd    →     Cd2+    +     2e-, Eo = -0.403 v
Reaction at cathode:     2MnO4-   +    2e-     →      MnO4^2-, Eo = 1.5 v

Net reaction: Cd  +   2MnO4-    ⇆     Cd2+     +     MnO4^2-

Net cell representation: Cd/Cd2+// MnO4-/MnO4^2-

Now, standard EMF of cell = E(o)cell = Er - El
                                                            = 1.5 - (-0.403)
                                                            = 1.903 v

Now, ΔGo = -nFE(o)cell
where n = number of electrons = 2
F = faraday's constant = 96500

ΔGo = - 2 X 96500 X 1.903
         = 367.27 kJ

Wind will blow from one direction to another.

The answer is 65.3312