For the cell-Tl s - Tl -0-001 M Cu -20-1 M - Cu s E cell -0-83 V at 298 K- the cell potential can be increased byA- Increasing - Cu -2-B- Increasing - Tl -C- Decreasing - Cu -2-D- Decreasing - T 1-

The correct options are A Increasing [Cu^+2] D Decreasing [Tl^+]Cell reaction is 2Tl_(s) + Cu^+2_(aq)⇌ 2Tl^+_(aq)+Cu_(s) Q_c = [Tl^+]^2/[Cu^+2] E=E^0 0.0591/nl ...

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Byju's Answer

Standard XII

Chemistry

Activation Energy

For the cell:...

Question

For the cell:
$T l_{(s)} | T l^{+} (0.001 M) | | C u^{+ 2} (0.1 M) | C u_{(s)}$
$E_{c e l l} = 0.83 V$ at 298 K, the cell potential can be increased by

Increasing

[C u^{+ 2}]

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Increasing

[T l^{+}]

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Decreasing

[C u^{+ 2}]

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Decreasing

[T l^{+}]

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Solution

The correct options are
A Increasing $[C u^{+ 2}]$
D Decreasing $[T l^{+}]$
Cell reaction is
$2 T l_{(s)} + C u_{(a q)}^{+ 2} ⇌ 2 T l_{(a q)}^{+} + C u_{(s)}$
$Q_{c} = \frac{[T l^{+}]^{2}}{[C u^{+ 2}]}$
$E = E^{0} - \frac{0.0591}{n} l o g Q_{c} a t 298 K$
$E = E^{0} - \frac{0.0591}{n} l o g (\frac{[T l^{+}]^{2}}{[C u^{+ 2}]})$

Suggest Corrections

Similar questions

For the cell:

${T l}_{(s)}$ / ${T l}^{+}$ (0.001M)// ${C u}^{+ 2}$ (0.1M)/ ${C u}_{(s)}$ :

$E_{c e l l}$ = 0.83V at 298K, The cell potential can be increased by

Q. For the cell:-

T l (s) | T l^{+} (0.001 M) | | C u^{2 +} (0.1 M) | C u (s); E_{c e l l} = 0.83 V

298 K

. The cell potential can be increased by :

Q. For the cell,

T l | T l^{+} (0.001 M) | | C u^{2 +} (0.1 M) | C u

, the

E_{c e l l}

25^{o}

C is 0.83V which can be increased by:

Q. For the cell:

T l_{(s)} | T l^{+} (0.001 M) | | C u^{2 +} (0.1 M) | C u_{(s)} :

E_{c e l l}

= 0.83 V at 298 K, The cell potential can be increased by

By how much will the potential of half cell cu+2/Cu change, if the solution is diluted to 100 times at 298 K?

(1) Increases by 59 mV

(2) Decreases by 59 mV

(3) Increases by 29.5 mv

(4) Decreases by 29.5 mV

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For the cell: Tl(s)|Tl+(0.001 M)||Cu+2(0.1 M)|Cu(s) Ecell=0.83 V at 298 K, the cell potential can be increased by

The correct options are A Increasing [Cu+2] D Decreasing [Tl+]Cell reaction is 2Tl(s)+Cu+2(aq)⇌2Tl+(aq)+Cu(s) Qc=[Tl+]2[Cu+2] E=E0−0.0591nlogQcat298K E=E0−0.0591nlog([Tl+]2[Cu+2])

For the cell:
$T l_{(s)} | T l^{+} (0.001 M) | | C u^{+ 2} (0.1 M) | C u_{(s)}$
$E_{c e l l} = 0.83 V$ at 298 K, the cell potential can be increased by