Question
Asked Nov 26, 2019
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Compute the equilibrium constant at 25 ∘C for the reaction between Sn2+(aq) and Cd(s), which form Sn(s) and Cd2+(aq).

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Expert Answer

Step 1

The given reaction is as follows:

Cd(s) +Sn2(aq) ->Cd2(aq) +Sn(s)
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Cd(s) +Sn2(aq) ->Cd2(aq) +Sn(s)

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Step 2

The calculation of the cell poten...

At anode:
Oxidation: Cd (s) -» Cd22(aq) +2e
,E= +0.403 V
*
cell
At cathode:
Reduction: Sn2(aq)+2e
Sn(s)E -0.14V
'cell
EE
+Eo
anode
cell
cathode
=-0.14 V+0.403 V
=0.263 V
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At anode: Oxidation: Cd (s) -» Cd22(aq) +2e ,E= +0.403 V * cell At cathode: Reduction: Sn2(aq)+2e Sn(s)E -0.14V 'cell EE +Eo anode cell cathode =-0.14 V+0.403 V =0.263 V

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