Question

♡ Search this course Faded 0 x cell potentials given below. Standard Reduction Potentials (Volts) at 25 °C 12() + 2 6 2 I'(aq) 0.535 02(g) + 4 Hz0+ (aq) + 4 € -6 H,00 1.229 2 H2O(l) + 2 € H2(g) + 2 OH"(aq) -0.828 Sc3(aq) + 3 e Sc(s) -2.080 Half-reaction at anode: Half-reaction at cathode:

Answer 1

Solution :

Here we can see Iodide ( I) shows a positive reduction potential and Scandium (Sc) shows a negative reduction potential and according to electrochemical seies, half reaction showing positive reduction potential shows reduction and half reaction showing negative reduction potential shows oxidation.

Means the half reaction which has positive reduction potential will be reduction half reaction and the half reaction which has negative reduction potential will be oxidation half reaction.

Hence, among all half reactions, Sc is oxidised and I is reduced and we know oxidation occurs at Anode and reduction occurs at Cathode.

Hence, overall half reactions are:

At Anode : Sc​​​​​3+ (s) + 3e-  -------> Sc (aq)

At Cathode : I​​​​​​2 (aq) + 2e- ------> 2I-  (s)

These are the half reactions at Anode and Cathode.

B) Decomposition potential (Vd)

It can be calculated by using the formula:

Vd = -nF∆G

Where, n = no of electrons ( 2 in this reaction )

F = 96500

∆G can be calculated by using the formula:

∆G = -nFE°cell

Where , E°cell = Standard reduction potential of RHS electrode - Standard reduction potential of LHS electrode

Hence, E°cell = 0.535 - ( - 2.080 )

Or, E°cell = 2.615 V

Hence, ∆G = - 2 × 96500 × 2.615

Or, ∆G = -505 KJ

And, Vd = -2 × 96500 × ( -505 )

Or, Vd = 974.6×108