For the following I M solutions, predict the product formed at the anode and cat
ID: 1066900 • Letter: F
Question
For the following I M solutions, predict the product formed at the anode and cathode and write balanced half-reactions for the processes that occur at each electrode [At pH = 7, E = -1.4 V for the oxidation of water at a Pt electrode (overvoltage correction applied): 2H_2O(l) - O_2(g) + 4H^+(aq) + 4e^-; while at pH = 7, E = -0.4 V for the reduction of water at a Pt electrode: 2H_2O(l) + 2e^- rightarrow H_2(g) + 2OH^- (aq). Consult your lecture textbook for a table of standard reduction potentials]. KF_(aq) NiBr_2(aq)Explanation / Answer
E0K+/K = -2.931 V
E0F2/F- = +2.866 V
E0Br2/Br- = + 1.0873 V
E0Ni2+/Ni = - 0.257
E0O2/H2O = + 1.4 V
E0H2O/H2 = - 0.4 V
(a)
KF (Aq.) ------------> K+ (aq.) + F- (aq.)
H2O (l) <------------> H+ (aq.) + OH- (aq.)
Reduction potential of H2O is more than K+ so, it undergoes reduction to produce H2 gas at cathode.
Oxidation potential of H2O is more than F-, so it under goes oxidation to prodeuce O2 at anode.
at anode:
2 H2O (l) ----> O2 (g) + 4 H+ (aq.) + 4 e
at cathode:
2 H2O (l) + 4 e- .------> H2 (g) + 2 OH- (aq.)
(b) NiBr2 (aq.) -------------> Ni2+ (aq.) + 2 Br- (Aq.)
H2O (l) <------------> H+ (aq.) + OH- (aq.)
Reduction potential of Ni2+ is more than H2O, so it undergoes reduction to form Ni solid at cathode.
Oxidation potential of Br- Is more than H2O, so it undergoes oxidation at anode to produce Br2(l) at anode.
At anode:
2 Br- (aq.) -------------> Br2 (l) + 2 e
at cathode:
Ni2+ (aq.) + 2 e ------------> Ni (s)