$Cu_{(s)} | Cu^{+2}(aq, 10^{-3} M) || Ag^{+}(aq, 10^{-5} M) | Ag_{(s)}$
If $E^{o}_{Cu^{+2}/Cu} = +0.34 \ V$
$E^{o}_{Ag^{+}/Ag} = +0.80 \ V$
$E_{cell}$ will be

What is the change in the reduction potential of a hydrogen electrode when the $pH$ of the initial solution changes from $0$ to $7$ by neutralization?

What will be the electromotive force of the following cell $Fe|Fe^{+2}\,(0.2\, M)\,||\, Au^{+3}\,(0.02\, M)\,|Au$ (in $,V$)?
Given: $E^o_{Fe^{+2}/Fe} = -0.44\,V$ and $E^o_{Au^{+3}/Au} = +1.50\,V$.

In the electrochemical cell $:$
$Zn \,|\,ZnSO_4\,(0.01\,M)\,||\,CuSO_4\,(1.0\,M)\,|\,Cu$
the $emf$ of this Daniell cell is $E_1.$ When the concentration of $ZnSO_4$ is changed to $1.0\,M$ and that of $CuSO_4$ changed to $0.01\,M,$ the $emf$ changes to $E_2.$ From the followings,which one is the relationship between $E_1$ and $E_2$ $?$ (Given,$RT/F = 0.059$)

Give the $E_{cell}$ equation for the general reaction: $aA + bB \to cC + dD$.

For the cell reaction,$Zn_{(s)} + 2 Ag_{(aq)}^{+} \longrightarrow Zn_{(aq)}^{2+} + 2 Ag_{(s)}$,the cell potential is less than $E^{\circ}_{cell}$ by $0.0592 \ V$ at $298 \ K$ when: