In a cell reaction- Cus-2Ag-aq- Cu2-aq-2Ags- Eocell-0-46V- If the concentration of Cu2- ions is doubled then Eocell will be-

For a cell reaction Cu(s) + 2Ag⁺(aq) → Cu ²⁺+ 2 Ag⁺ (aq) the standard cell potential is 0.46V . By doubling the concentration of Copper(II) ions, The standard cell potential will get:-

Q. E.M.F. of the cell reaction,

2 A g^{+} + C u ⟶ 2 A g + C u^{2 +}

is 0.46 V. If

E_{{C u}^{2 +} / C u}

is 0.34 V,

E_{{A g}^{+} / A g}

is:

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In a cell reaction; Cu(s)+2Ag+(aq)→Cu2+(aq)+2Ag(s), Eocell=+0.46V. If the concentration of Cu2+ ions is doubled then Eocell will be:

The correct option is D unchangedWe know that,E0cell=E0cathode−E0anodeThe equation shows that E0cell doesn't depend on the concentration of the ions present in it and hence the value will remain unchanged.Hence option (D) is correct

In a cell reaction; $C u_{(s)} + 2 A g_{(a q)}^{+} \to C u_{(a q)}^{2 +} + 2 A g_{(s)}$ , $E_{c e l l}^{o} = + 0.46$ V. If the concentration of $C u^{2 +}$ ions is doubled then $E_{c e l l}^{o}$ will be:

The correct option is D unchanged
We know that,
$E_{c e l l}^{0}$ = $E_{c a t h o d e}^{0} - E_{a n o d e}^{0}$
The equation shows that $E_{c e l l}^{0}$ doesn't depend on the concentration of the ions present in it and hence the value will remain unchanged.
Hence option (D) is correct