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Calculate E("cell")^(@) for the followin...

Calculate `E_("cell")^(@)` for the following reaction at 298K : `2Al(s)+3Cu^(2+)(0.01M)rarr 2Al^(3+)(0.01M)+3Cu(s)`
[GIven `E_(cell)=1.98V`]

Text Solution

Verified by Experts

For the cell reaction, n = 3
Applying Nernst's equation, we get
`E_("cell")=E_("cell")^(@)-(0.059)/(3)log.([Al^(3+)]^(2))/([Cu^(2+)]^(3))`
`1.98=E_("cell")^(@)-(0.059)/(3)log.((0.01)^(2))/((0.01)^(3))`
`therefore" "E_("cell")^(@)=2.02V`
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Knowledge Check

  • At 25^(@)C temperature, E_(Zn^(2+)|Zn)^(@)=-0.76V and E_(Ag^(2+)|Ag)^(@)=+0.80V. If the E_("cell")^(@) for a galvanic cell formed by combining the two half - cells, Zn^(2+)(1M)|Zn and Cu^(2+)(1M)|Cu is 1.1V , then E_("cell")^(@) for a galvanic cell formed by combining the two half - cells, Cu^(2+)(1M)|Cu and Ag^(+)(1M)|Ag, will be -

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