Find the EMF of the following cell : $$\ce{Pb(s)}, \ce{PbSO_4}|\ce{SO_4^{2-}}(\pu{0.100M})||\ce{Pb^{2+}}(\pu{0.004M})|\ce{Pb(s)}$$ Given: $E^0_{\ce{PbSO_4|Pb,SO_4^{2-}}}=\pu{-0.359V}$ and $E^0_{\ce{Pb^{2+}|Pb}}=\pu{-0.126V}$

I first found the $\ce{Pb^{2+}}$ concentration in the oxidation half cell using the sulphate ion concentration and the solubility product of lead sulpahte ($2.53\times10^{-8}$) and found the concentration to be $(2.53\times10^{-7})$ And accordingly $E^0_{\text{cell}}=0.359-0.126=0.233$ Then using the Nernst equation, $$E_{\text{cell}}=E^0_{\text{cell}}-\frac{RT}{nF}\ln\frac{[\text{Products}]}{[\text{Reactants}]}$$ And substituting, $n=2$, $[\text{Products}]=2.53\times10^{-7}$, $[\text{Reactants}]=0.004$, $T=\pu{298K}$, $R=8.314JK^{-1}mol^{-1}$, and$F=96500C$. Hence, I got $E_{\text{cell}}=\pu{0.357V}$

Whereas, the answer given in my book is $E_{\text{cell}}=\pu{0.133V}$.

So, is my answer correct or have i misunderstood something?

Find the EMF of the following cell : $$\ce{Pb(s)}, \ce{PbSO_4}|\ce{SO_4^{2-}}(\pu{0.100M})||\ce{Pb^{2+}}(\pu{0.004M})|\ce{Pb(s)}$$ Given: $E^0_{\ce{PbSO_4|Pb,SO_4^{2-}}}=\pu{-0.359V}$ and $E^0_{\ce{Pb^{2+}|Pb}}=\pu{-0.126V}$

I first found the $\ce{Pb^{2+}}$ concentration in the oxidation half cell using the sulphate ion concentration and the solubility product of lead sulpahte ($2.53\times10^{-8}$) and found the concentration to be $(2.53\times10^{-7})$ And accordingly $E^0_{\text{cell}}=0.359-0.126=0.233$ Then using the Nernst equation, $$E_{\text{cell}}=E^0_{\text{cell}}-\frac{RT}{nF}\ln\frac{[\text{Products}]}{[\text{Reactants}]}$$ And substituting, $n=2$, $[\text{Products}]=2.53\times10^{-7}$, $[\text{Reactants}]=0.004$, $T=\pu{298K}$, $R=\pu{8.314JK^{-1}mol^{-1}}$, and $F=\pu{96500C}$. Hence, I got $E_{\text{cell}}=\pu{0.357V}$

Whereas, the answer given in my book is $E_{\text{cell}}=\pu{0.133V}$.

So, is my answer correct or have i misunderstood something?

Find the EMF of the following cell : $$\ce{Pb(s)}, \ce{PbSO_4}|\ce{SO_4^{2-}}(\pu{0.100M})||\ce{Pb^{2+}}(\pu{0.004M})|\ce{Pb(s)}$$ Given: $E^0_{\ce{PbSO_4|Pb,SO_4^{2-}}}=\pu{-0.359V}$ and $E^0_{\ce{Pb^{2+}|Pb}}=\pu{-0.126V}$

I first found the $\ce{Pb^{2+}}$ concentration in the oxidation half cell using the sulphate ion concentration and the solubility product of lead sulpahte ($2.53\times10^{-8}$) and found the concentration to be $(2.53\times10^{-7})$ And accordingly $E^0_{\text{cell}}=0.359-0.126=0.233$ Then using the Nernst equation, $$E_{\text{cell}}=E^0_{\text{cell}}-\frac{RT}{nF}\ln\frac{[\text{Products}]}{[\text{Reactants}]}$$ And substituting, $n=2$, $[\text{Products}]=2.53\times10^{-7}$, $[\text{Reactants}]=0.004$, $T=\pu{298K}$, $R=8.314$, and$F=96500$. Hence, I got $E_{\text{cell}}=\pu{0.357V}$

Whereas, the answer given in my book is $E_{\text{cell}}=\pu{0.133V}$.

So, is my answer correct or have i misunderstood something?

Find the EMF of the following cell : $$\ce{Pb(s)}, \ce{PbSO_4}|\ce{SO_4^{2-}}(\pu{0.100M})||\ce{Pb^{2+}}(\pu{0.004M})|\ce{Pb(s)}$$ Given: $E^0_{\ce{PbSO_4|Pb,SO_4^{2-}}}=\pu{-0.359V}$ and $E^0_{\ce{Pb^{2+}|Pb}}=\pu{-0.126V}$

I first found the $\ce{Pb^{2+}}$ concentration in the oxidation half cell using the sulphate ion concentration and the solubility product of lead sulpahte ($2.53\times10^{-8}$) and found the concentration to be $(2.53\times10^{-7})$ And accordingly $E^0_{\text{cell}}=0.359-0.126=0.233$ Then using the Nernst equation, $$E_{\text{cell}}=E^0_{\text{cell}}-\frac{RT}{nF}\ln\frac{[\text{Products}]}{[\text{Reactants}]}$$ And substituting, $n=2$, $[\text{Products}]=2.53\times10^{-7}$, $[\text{Reactants}]=0.004$, $T=\pu{298K}$, $R=8.314JK^{-1}mol^{-1}$, and$F=96500C$. Hence, I got $E_{\text{cell}}=\pu{0.357V}$

Whereas, the answer given in my book is $E_{\text{cell}}=\pu{0.133V}$.

So, is my answer correct or have i misunderstood something?

A question in my book asks us to find the EMF of the following cell : $$Pb(s),PbSO_4|SO_4^{-2}(0.100M)||Pb^{+2}(0.004M)|Pb(s)$$ Given: $$E^0_{PbSO_4|Pb,SO_4^{-2}}=-0.359V$$ $$E^0_{Pb^{+2}|Pb}=-0.126V$$

I first found the $Pb^{+2}$ concentration in the oxidation half cell using the Sulphate ion concentration and the solubility product of Lead Sulpahte($2.53\times10^{-8}$) and found the concentration to be ($2.53\times10^{-7}$) And accordingly $E^0_{cell}=0.359-0.126=0.233$ Then using the Nernst equation, $$E_{cell}=E^0_{cell}-\frac{RT}{nF}\ln\frac{[Products]}{[Reactants]}$$ And substituting, $$n=2$$ $$[Products]=2.53\times10^{-7}$$ $$[Reactants]=0.004$$ $$T=298K$$ $$R=8.314$$ $$F=96500$$ I get $E_{cell}=0.357V$

Whereas, the answer given in my book is $E_{cell}=0.133V$

So, is my answer correct or have i misunderstood something?

Find the EMF of the following cell : $$\ce{Pb(s)}, \ce{PbSO_4}|\ce{SO_4^{2-}}(\pu{0.100M})||\ce{Pb^{2+}}(\pu{0.004M})|\ce{Pb(s)}$$ Given: $E^0_{\ce{PbSO_4|Pb,SO_4^{2-}}}=\pu{-0.359V}$ and $E^0_{\ce{Pb^{2+}|Pb}}=\pu{-0.126V}$

I first found the $\ce{Pb^{2+}}$ concentration in the oxidation half cell using the sulphate ion concentration and the solubility product of lead sulpahte ($2.53\times10^{-8}$) and found the concentration to be $(2.53\times10^{-7})$ And accordingly $E^0_{\text{cell}}=0.359-0.126=0.233$ Then using the Nernst equation, $$E_{\text{cell}}=E^0_{\text{cell}}-\frac{RT}{nF}\ln\frac{[\text{Products}]}{[\text{Reactants}]}$$ And substituting, $n=2$, $[\text{Products}]=2.53\times10^{-7}$, $[\text{Reactants}]=0.004$, $T=\pu{298K}$, $R=8.314$, and$F=96500$. Hence, I got $E_{\text{cell}}=\pu{0.357V}$

Whereas, the answer given in my book is $E_{\text{cell}}=\pu{0.133V}$.

So, is my answer correct or have i misunderstood something?