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Calculate the reduction potential for th...

Calculate the reduction potential for the following half cell reaction at 298 K.
`Ag^(+)(aq)+e^(-)toAg(s)`
`"Given that" [Ag^(+)]=0.1 M and E^(@)=+0.80 V`

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To calculate the reduction potential for the half-cell reaction \( \text{Ag}^+(aq) + e^- \rightarrow \text{Ag}(s) \) at 298 K, we will use the Nernst equation: \[ E = E^\circ - \frac{0.0591}{n} \log Q \] ### Step-by-Step Solution: ...
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DINESH PUBLICATION-ELECTROCHEMISTRY-Example
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  10. Calculate e.m.f. of the cell, Zn//Zn^(2+)(aq) (0.01 M) ||Cd^(2+) (0.1...

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  11. Calculate the e.m.f. of the cell in which the redox reaction is : Mg...

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  12. Calculate the e.m.f of the cell Mg(s)//Mg^(2+)(0.1 M)||Cu^(2+)(1.0xx...

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  13. A voltaic cell is set up at 25^(@)C with following half cells : Ag^(+)...

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  16. A voltaic cell is set up at 25^(@)Cwith the following half cells : A...

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  17. For the electrochemical cell, Mg(s)|Mg^(2+)(aq,1 M)||Cu^(2+)(aq.1 M) C...

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  18. A cell contains two hydrogen electrodes. The negative electrode is in ...

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  19. Calculate the emf of the cell : Pb(s)|Pb(NO(3))(2)(M(1))||HCl(M(2))|H...

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  20. The obseved emf of the cell Pt//H(2)(g, 1" atm")| H^(+)(3xx10^(-4) M...

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