Question

calculate the electrode potential of a copper wire dipped in 0.1 M CuSO4 solution at 25°C. the standard potential of copper is 0.34 volt.​

Answer 1

Answer :

• Electrode potential of copper wire = 0.31045 volt.

Step-by-step explanation :

Given that,

• Standard electrode potential of copper = 0.34 volt.

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The electrode reaction written as reduction reaction is as follows :-

$\rm{Cu^{2+}+2\:e^-\longrightarrow{}Cu}$

Now, applying Nernst eqn. we get,

$\implies\rm{E=E^{\circ}+\dfrac{0.0591}{n}log_{10}[ion]}$

Putting the values, we get,

$\implies\rm{E=0.34+\dfrac{0.0591}{2}log_{10}[0.1]}$

$\implies\rm{E=0.34+0.02955\times{}(-1)}$

$\implies\rm{E=0.34-0.02955}$

$\implies\bold{E=0.31045\:volt.}$

Answer 2

Answer:

Calcalate the electrode potential of a copper wire dipped in 0.1 M CuSO4 solution at 25∘C. The standard electrode potential of copper is 0.34 volt. =0.34-0.05912log10.1=0.34-0.02955=0.31045volt.