Pd/Pd2+ (aq, 0.00010M)/ Br2 (l), Br-(aq, 0.00010M)/Pt(s) What is the Ecell?

Pd/Pd²⁺ (aq, 0.00010M)/ Br₂ (l), Br^-(aq, 0.00010M)/Pt(s)

What is the E_cell?

Expert Solution

Pb(s) ---------------> Pb^2+ (aq) + 2e^- E0 = 0.13v

Br2(l) + 2e^- ---------> 2Br^- (aq) E0 = 1.07v

---------------------------------------------------------------------------------------

Pb(s) + Br2(l) ------------> Pb^2+ (aq) + 2Br^-(aq) E0cell = 1.20v

n = 2

Ecell = E0cell- 0.0592/n logQ

= 1.20-0.0592/2 log[Br^-]^2[Pb^2+]

= 1.20-0.0296log(0.0001)^2*(0.0001)

= 1.20-0.0296log1*10^-12

= 1.20-0.0296*-12log10

= 1.20+0.3552

= 1.5552v >>>>answer

queen_honey_blossom answered 2 years ago

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