WRITING OF CELL REACTION

WRITING OF CELL REACTION

(I) The cell reaction corresponding to the cell diagram is written on the assumption that the right-hand electrode is the cathode (+) and the left-hand electrode is the anode (-). 

(II) In all the voltaic cells the electrons flow spontaneously from left to right that is from the negative electrode to the positive electrode through an external circuit. This implies that electrons are released at the left-hand side electrode (anode) and consumed at the right-hand side electrode (cathode). In other words oxidation, half-reaction takes place at left-hand side electrode and reduction half-reaction at right-hand side electrode. 

(III) The overall cell reaction is the redox reaction which is the sum of oxidation half-reaction at anode and reduction half-reaction at the cathode. 

(IV) While adding the half-reactions the electrons must cancel. For this purpose, it may be necessary to multiply one or both the half-reactions by a suitable numerical factor(s). 

(V) No electrons appear in the overall cell reaction because all the electrons lost by the species undergoing oxidation are gained by the species that undergoes reduction. 

(VI) It is important to note that the individual half reactions may be written with one or more electrons. 

For example, 

half-reaction for H₂ gas whether written as

2H⁺_(aq) + 2e^-→ H_2(g) or as H⁺_(aq) + e^- → 1/2 H_2(g) makes no difference. However, in writing the overall cell reaction, obviously, the electrons must be balanced. 

The following examples will illustrate the above arguments: 

(i) Consider the cell, 

Pb_(s) | Pb²⁺(1M) || Ag⁺(1M) | Ag 

The oxidation half reaction at anode is, Pb_(s) → Pb²⁺(1M) + 2e^- 

The reduction half-reaction at the cathode is, Ag⁺(1M) + e^- → Ag_(s) 

The overall cell reaction is obtained by multiplying the reduction half-reaction by 2 to balance the electrons and then adding to the oxidation half-reaction. 

 

Pb(s) → Pb²⁺ (1M) + 2e^- (oxidation half reaction) 

2Ag⁺ (1M) + 2e^- → 2 Ag (s) (reduction half reaction) 

Pb(s) + 2Ag + (1M) → Pb²⁺ (1M) + 2 Ag (overall cell reaction) 

(ii) Consider the cell 

Pt (s) | H₂ (g, 1atm) | H⁺(1M) || Cr³⁺(1M) | Cr(s) 

The oxidation half-reaction at anode is H₂ (g, 1atm) → 2H⁺(1M) + 2e^- 

The reduction half-reaction at cathode is Cr³⁺ (1M) + 3e^- → Cr(s) 

To write the overall cell reaction, 

oxidation half-reaction is multiplied by 3 and reaction half-reaction is multiplied by 2 and then the modified half-reactions are added together. 

3H₂ (g, 1atm) → 6H⁺ (1M) + 6e^- (oxidation half reaction) 

2Cr³⁺ (1M) + 6e^- → 2Cr(s) (reduction half reaction) 

3H₂ (g, 1atm) + 2Cr³⁺ (1M) → 6H⁺ (1M) + 2Cr(s) (overall cell reaction)