Question

A copper wire is dipped ion AgNO3 solution kept in a beaker A and a silver wire is dipped in a solution of copper sulphate kept in a beaker B.If the standard electrode potential for Cu^+2 + 2e------Cu is 0.34V and for the Ag^+ + e--------Ag is 0.80V predict in which beaker the ions present will get reduced?​

Answer 1

Solution :

The probable reaction in beaker A will be <br>

<br>

for this reaction

i.e., it is +ve. Hence,

ions are reduced to Ag.

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Answer 2

Answer:

The probable reaction in beaker A will be

Cu + 2AgNO3 → Cu(NO3)2 + 2Ag,

i.e.,

Cu + (2Ag+) → (Cu2+) + 2Ag

Ecell for this reaction =−0.34+0.80=+0.46V i.e., it is +ve. Hence, Ag+ ions are reduced to Ag.

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