Question

In: Chemistry

Consider the cell described below at 291 K: Sn | Sn2+ (0.753 M) || Pb2+ (0.921...

Consider the cell described below at 291 K:

Sn | Sn²⁺ (0.753 M) || Pb²⁺ (0.921 M) | Pb

Given E^o_Pb²⁺→Pb = -0.131 V, E^o_Sn²⁺→Sn = -0.143 V. Calculate the cell potential after the reaction has operated long enough for the Sn²⁺ to have changed by 0.325 mol/L.

Expert Solution

anode reaction: oxidation takes place

Sn(s) -------------------------> Sn+2 (aq) + 2e- , E⁰_Sn+2/Sn = - 0.143 V

cathode reaction : reduction takes palce

Pb+2(aq) + 2e- -----------------------------> Pb(s) , E⁰_Pb+2/Pb = -0.44V

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

net reaction: Sn(s) +Pb+2(aq) -------------------------> Sn+2 (aq) + Pb(s)

E⁰_cell= E⁰_cathode- E⁰_anode

E⁰_cell= E⁰_Pb+2/Pb - E⁰_Sn+2/Sn

= -0.131 - (-0.143)

= 0.012V

nernest equation

Ecell = E⁰_cell -2.303RT/nF* log [Sn+2]/[Pb+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 =2

2.303RT/F= 0.0591

Ecell = E⁰_cell -(0.0591/n)* log [Sn+2]/[Pb+2]

Ecell = 0.012 - (0.059 x1/2) *log [0.753/0.921]

Ecell = 0.0145 V

cell potential =Ecell = 0.0146 V

queen_honey_blossom answered 11 hours ago

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