how would oh- in an ammonia solution compare With oh- in a sodium hydorxide solution if similar consentration
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Answer:
E cell = +1.95 V
Explanation:
At Anode : Oxidation reaction takes place
At Cathode : Reduction reaction takes place
The reaction with lower value of reduction potential will undergo Oxidation
E = -1.18 V
This equation undergo oxidation reaction and become:
Anode(Oxidation-Half) :
E = +1.18 V
Cathode(Reduction-Half) :
E =+0.77 V
To balance the reaction multiply reduction-Half with 2.We get :
Note that E is intensive property , do not multiply E of oxidation-half with 2
Ecell = 0.77 -(-1.18)
E = +1.95 V