The electrochemical cell shown below is a concentration cell M|M^ 2(Mx2)||M^ 2(0.001m)|M The emf of cell depends on the difference in concentration M^ 2 ion at two electrode.The emf of cell at 298k is 0.059v... Find solubility product of Mx2 at 298k???

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Naveen Kumar · Answer

In this question the cell notation shown is not clear.................Please write clearly and specify the concentrations and all other things given in the queston.

Ashish Papanai · Answer

Use Nernst Equation Ecell = E0cell - (0.059/n)*LogK Use the formula :- ∆G = -nEF Answer will be -11.4 (d)

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The electrochemical cell shown below is a concentration cell M|M^ 2(Mx2)||M^ 2(0.001m)|M The emf of cell depends on the difference in concentration M^ 2 ion at two electrode.The emf of cell at 298k is 0.059v... Find solubility product of Mx2 at 298k???

The electrochemical cell shown below is a concentration cell M|M^ 2(Mx2)||M^ 2(0.001m)|M
The emf of cell depends on the difference in concentration M^ 2 ion at two electrode.The emf of cell at 298k is 0.059v...
Find solubility product of Mx2 at 298k???

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The electrochemical cell shown below is a concentration cell M|M^ 2(Mx2)||M^ 2(0.001m)|M The emf of cell depends on the difference in concentration M^ 2 ion at two electrode.The emf of cell at 298k is 0.059v... Find solubility product of Mx2 at 298k???

The electrochemical cell shown below is a concentration cell M|M^ 2(Mx2)||M^ 2(0.001m)|M The emf of cell depends on the difference in concentration M^ 2 ion at two electrode.The emf of cell at 298k is 0.059v...Find solubility product of Mx2 at 298k???

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The electrochemical cell shown below is a concentration cell M|M^ 2(Mx2)||M^ 2(0.001m)|M
The emf of cell depends on the difference in concentration M^ 2 ion at two electrode.The emf of cell at 298k is 0.059v...
Find solubility product of Mx2 at 298k???