Question

In: Chemistry

An electrochemical cell is based on the following two half-reactions: Ox: Pb(s)?Pb2+(aq, 0.21M )+2e? Red: MnO?4(aq,...

An electrochemical cell is based on the following two half-reactions:
Ox: Pb(s)?Pb2+(aq, 0.21M )+2e?
Red: MnO?4(aq, 1.25M )+4H+(aq, 2.5M )+3e??
MnO2(s)+2H2O(l). Compute the cell potential at 25 ?C.

Expert Solution

Oxidation : Pb(s)---------------> Pb2+ (0.21M )+2e - , E⁰_Pb+2/Pb = -0.13 V
Reduction : MnO^-4(aq, 1.25M )+4H+(aq, 2.5M )+3e- ---------------->MnO2(s)+2H2O(l). , E⁰_MnO4-/MnO2 = 1.70V
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net reaction

3Pb(s) + 2MnO4- (1.25M) + 8 H+ (2.5M) ------------------------> 3 Pb+2 (0.21 M) + 2MnO2(s) + 4H2O(l)

E⁰_cell= E⁰_cathode- E⁰_anode

E⁰_cell= E⁰_MnO4-/_MnO2- E⁰_Pb+2/Pb

= 1.70 -(-0.13)

= 1.83 V

now nernest equation

Ecell = E⁰_cell -RT/nF* log [Mg+2]/[Ni+2]

Here R= universal gas constant 8.314 J/K mol

T = absolute temperature =25(0C)= 298k

F= faraday = 96500 Coloumb/mol

n = no of moles of electrons are transfered =6

RT/F= 0.0591

Ecell = E⁰_cell -0.0591/n* log [Pb+2]^3/[MnO4-]^2[H+]^8

Ecell = 1.83 -0.0591/6* log [0.21]^3 / [1.25]^2[2.5]^8

= 1.83 - 0.0591/6* log [0.21]^3 / [1.25]^2[2.5]^8

Ecell = 0.928V ( by simplipying whole value)

cell potential = Ecell = 0.928V

queen_honey_blossom answered 1 year ago

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