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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`A. 0.741 VB. 0.80 VC. `-0.80`VD. `-0.741`V |
Answer» Correct Answer - A `Ag^(+) + e^(-) to Ag` `E_"cell"=E_"cell"^@-0.0591/1 "log" 1/([Ag^+])` `E_"cell"=0.80-0.0591/1 "log" 1/0.1` =0.80-0.591=0.741 V |
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