    Question

# Calculate the voltage E in V, of the cellAg(s)AgIO3(s)|Ag+(xM),HIO3(0.300M)|∣∣Zn2+(0.175M)∣∣Zn(s)if KSP=3.02×10−8 for AgIO3(s) and Ka=0.162 for HIO3,E∘(Zn2+/Zn)=−0.76V,E∘(Ag/Ag+=−0.8V)

A
0.45
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B
0.32
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C
1.88
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D
1.45
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Solution

## The correct option is A 0.45E0cell=E0R−E0L=−0.76−(−0.8)=0.04VHIO3H+IO−3Initial concentration (M)0.30000Final concentration (M)0.300−xxxThe equilibrium constant expression is K=[H+][IO−3][HIO3]0.162=x×x0.3−x6.173x2+x−0.3=0This is quadratic equation with solutions -0.31 and 0.154. As concentration cannot be negative, the value -0.31 is discarded. [IO−3]=x=0.154MKsp=[Ag+][IO−3]3.02×10−8=[Ag+]×0.154[Ag+]=19.6×10−8=1.96×10−7MZn2++2Ag→Zn+2Ag+Ecell=E0cell−0.0592nlog[Ag+]2[Zn2+]Ecell=0.04−0.05922log(1.96×10−7)20.175Ecell=0.04+0.375=0.415V  Suggest Corrections  0      Similar questions  Related Videos   Nernst Equation
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