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  • Given that, <img alt="\mathrm{E_{Ag^+/Ag}^{0} = 0.8V\: and \: E_{ Zn/Zn^2+}^{0}= 0.76V}" src="http://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7BE_%7BAg%5E&plus;/Ag%7D%5E%7B0%7D%20%3D%2

Given that, \mathrm{E_{Ag^+/Ag}^{0} = 0.8V\: and \: E_{ Zn/Zn^2+}^{0}= 0.76V} .
What will be the EMF of the cell made by connecting the given electrodes.

Option: 1

0.04


Option: 2

1.56


Option: 3

-0.04


Option: 4

2.36


Answers (1)

best_answer

Here, Ag^+/Ag will be the cathode and Zn/Zn^{2+}will anode.

E_{cell}^{0}= \left ( E_{C}^{0}-E_{A}^{0} \right ) _{RP}
            = \left ( E_{C}^{0} \right ) _{RP}+\left ( E_{A}^{0} \right )_{OP}
            = 0.8+0.75=1.56V

Therefore, option(2) is correct

Posted by

seema garhwal

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