Question

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Answer 1

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Answer 2

Answer:

Explanation:

$\huge\boxed{\fcolorbox{red}{Yellow}{Given Reaction}}</p><p>$

$\sf\:Sn(S)|Sn^{+2}(0.05M)\:||\:H^{+1}(0.02)|H_2(g)(1\:bar)|Pt(s)Sn(S)∣Sn+2+(0.05M)∣∣H + +1(0.02)∣H2(g)(1bar)∣Pt(s)$

$\huge \bold\green{To \: find }$

Calculate the potential of the given cell reaction at 298 k. and

$\sf\:E\degree\:Sn^{+2}/Sn=-0.14E°Sn+2+ /Sn=−0.14$

Nernst \: Equation

$\sf\:E_{cell}=E\degree_{cell}+\dfrac{-2.303RT}{nF}\log\dfrac{[Prduct]}{[Reactant]} \\ \implies \: or \\ \sf\:E_{cell}=E\degree_{cell}-\dfrac{0.059}{n}\log\:QE cell=E° cell − n0.059logQ$

$E°cell \ \: \\ \sf\:E\degree_{cell=E\degree_{cathode}-E\degree_{anode}}$

$\textbf{ \huge { Solution }}$

WE HAVE REACTION

$\sf\:Sn(S)|Sn^{+2}(0.05M)\:||\:H^{+1}(0.02M)|H_2(g)(1\:bar)|Pt(s)Sn(S)∣Sn+2+ (0.05M)∣∣H+ +1(0.02M)∣H2(g)(1bar)∣Pt(s)$

$\mathtt{we \: know \: that}$

$\bf\:E\degree_{cell}=E\degree_{cathode}-E\degree_{anode}$

$\sf\:E\degree_{cell}=0-(-0.14)E° </p><p>[tex]\sf\:E\degree_{cell}=0-(-0.14)E° cell=0−(−0.14)$

$\sf\:E\degree_{cell}=0.14VE° </p><p>[tex]\sf\:E\degree_{cell}=0.14VE° cell</p><p>[tex]\sf\:E\degree_{cell}=0.14VE° cell </p><p>[tex]\sf\:E\degree_{cell}=0.14VE° cell =0.14V$

$\mathtt{At \: anode = }$

$\sf\:Sn\longrightarrow\:Sn^{+2}2e^{-1}Sn⟶Sn </p><p>[tex]\sf\:Sn\longrightarrow\:Sn^{+2}2e^{-1}Sn⟶Sn +2 +2e −1$

$\mathtt{at \: cathode}$

$\sf\:2H^{+}+2e^{-1}\longrightarrow\:H_22H </p><p>[tex]\sf\:2H^{+}+2e^{-1}\longrightarrow\:H_22H +2e </p><p>[tex]\sf\:2H^{+}+2e^{-1}\longrightarrow\:H_22H +2e +−1</p><p>[tex]\sf\:2H^{+}+2e^{-1}\longrightarrow\:H_22H +2e +−1 + ⟶H </p><p>[tex]\sf\:2H^{+}+2e^{-1}\longrightarrow\:H_22H +2e +−1 + ⟶H + 2</p><p>[tex]\sf\:2H^{+}+2e^{-1}\longrightarrow\:H_22H +2e +−1 + ⟶H + 2 + </p><p></p><p>[tex]\sf\:2H^{+}+2e^{-1}\longrightarrow\:H_22H +2e +−1 + ⟶H + 2 + \\ \\ Thus, n= 2$

$\mathtt{over \: all \: reaction : }$

$\sf\:Sn+2H^{+1}\longrightarrow\:H_2+Sn^{+2}Sn+2H </p><p>[tex]\sf\:Sn+2H^{+1}\longrightarrow\:H_2+Sn^{+2}Sn+2H +1</p><p>[tex]\sf\:Sn+2H^{+1}\longrightarrow\:H_2+Sn^{+2}Sn+2H +1+ ⟶H </p><p>[tex]\sf\:Sn+2H^{+1}\longrightarrow\:H_2+Sn^{+2}Sn+2H +1+ ⟶H + 2</p><p>[tex]\sf\:Sn+2H^{+1}\longrightarrow\:H_2+Sn^{+2}Sn+2H +1+ ⟶H + 2 +Sn </p><p>[tex]\sf\:Sn+2H^{+1}\longrightarrow\:H_2+Sn^{+2}Sn+2H +1+ ⟶H + 2 +Sn +2</p><p>[tex]\sf\:Sn+2H^{+1}\longrightarrow\:H_2+Sn^{+2}Sn+2H +1+ ⟶H + 2 +Sn +2$

$\mathtt{then : }$

$\sf\:Q=\dfrac{[Prduct]}{[Reactant]}$

$\sf\:Q=\dfrac{[Sn^{+2}]}{[H^{+1}]^2}$

$\green{\underline{\underline{\huge{\mathfrak\red{Given}}}}}$

$\sf\:[Sn^{+2}]=0.05M\:and\:[H^{+1}]=0.02M$

$\sf\implies\:Q=\dfrac{0.05}{(0.02)^2}$

$\sf\implies\:Q=\dfrac{5\times100\times100}{2\times100}$

$\implies\:Q=250$

$\mathtt{By \: Nernst \: Equation :}$

$\sf\:E_{cell}=E\degree_{cell}-\dfrac{0.09}{n}\log\:Q$

$\sf\implies\:E_{cell}=0.14-\dfrac{0.059}{2}\log\:250$

$\sf\implies\:E_{cell}=0.14-\dfrac{0.056}{2}[\log(25\times10)]$

$\sf\implies\:E_{cell}=0.14-0.03[\log25+\log10]$

$\sf\implies\:E_{cell}=0.14-0.03[2\times0.69-1]$

$\sf\implies\:E_{cell}=0.14-0.03[1.38-1]$

$\sf\implies\:E_{cell}=0.14-0.03[0.38]$

$\sf\implies\:E_{cell}=0.14-0.0114$

$\sf\implies\:E_{cell}=0.1286V$

$\mathtt{Therefore,the \: potential \: of \: the \: given \: cell \: is \: 0.1286V.}$