Question

tandard reduction half-cell potentials at 25 ∘ C Half-reaction E ∘ ( V ) Half-reaction E ∘ ( V ) A u 3+ (aq)+3 e − →Au(s) 1.50 F e 2+ (aq)+2 e − →Fe(s) − 0.45 A g + (aq)+ e − →Ag(s) 0.80 C r 3+ (aq)+ e − →C r 2+ (aq) − 0.50 F e 3+ (aq)+3 e − →F e 2+ (aq) 0.77 C r 3+ (aq)+3 e − →Cr(s) − 0.73 C u + (aq)+ e − →Cu(s) 0.52 Z n 2+ (aq)+2 e − →Zn(s) − 0.76 C u 2+ (aq)+2 e − →Cu(s) 0.34 M n 2+ (aq)+2 e − →Mn(s) − 1.18 2 H + (aq)+2 e − → H 2 (g) 0.00 A l 3+ (aq)+3 e − →Al(s) − 1.66 F e 3+ (aq)+3 e − →Fe(s) − 0.036 M g 2+ (aq)+2 e − →Mg(s) − 2.37 P b 2+ (aq)+2 e − →Pb(s) − 0.13 N a + (aq)+ e − →Na(s) − 2.71 S n 2+ (aq)+2 e − →Sn(s) − 0.14 C a 2+ (aq)+2 e − →Ca(s) − 2.76 N i 2+ (aq)+2 e − →Ni(s) − 0.23 B a 2+ (aq)+2 e − →Ba(s) − 2.90 C o 2+ (aq)+2 e − →Co(s) − 0.28 K + (aq)+ e − →K(s) − 2.92 C d 2+ (aq)+2 e − →Cd(s) − 0.40 L i + (aq)+ e − →Li(s) − 3.04 You may want to reference (Pages 749 - 752) Section 18.5 while completing this problem. Part A Use the tabulated electrode potentials to calculate K for the oxidation of nickel by H + : Ni(s)+2 H + (aq)→N i 2+ (aq)+ H 2 (g) Express your answer using two significant figures.

Answer 1

Ni(s)+2 H + (aq)→N i 2+ (aq)+ H2

For this reaction number of electron involved n=2.

E⁰ cell= E⁰ cathode^- E⁰  anode

E⁰ anode= -0.23v(Ni half cell)

E⁰ cathode= 0.00v(H2half cell)thus, E⁰ cell=-0.23v

Now, as perNernst equation, E0cell=0.0591/n+log₁₀(K)

-0.23=0.0591÷2 +log(k)

Thus, K=1.684×10^-8