The standard oxidation potentials of iron and silver are +0.41 V and -0.80 V, respectively. The e.m.f. for cell reaction 2 Ag ( aq )++ Fe ( s ) longrightarrow 2 Ag ( s )+ Fe ( aq )2+ is V

Question

The standard oxidation potentials of iron and silver are +0.41 V and -0.80 V, respectively. The e.m.f. for cell reaction 2 Ag ( aq )++ Fe ( s ) longrightarrow 2 Ag ( s )+ Fe ( aq )2+ is V (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Reduction potential of Fe and Ag electrodes are Fe ( text aq )2++2 e - longrightarrow Fe ( s ), E °=-0.41 V Ag ( aq )++ e - longrightarrow Ag ( s ) ; E =+0.80 V <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/chemistry/e3af157173b4fe230201838a4062c984-.png" /> E text cell °= E text cathode °- E text anode °=+0.80-(-0.41) =1.21 V

Q. The standard oxidation potentials of iron and silver are $+ 0.41 V$ and $- 0.80 V$ , respectively. The e.m.f. for cell reaction
$2 A g_{(a q)}^{+} + F e_{(s)} ⟶ 2 A g_{(s)} + F e_{(a q)}^{2 +}$ is ________ $V$

Reduction potential of Fe and Ag electrodes are
$F e_{(aq)}^{2 +} + 2 e^{-} ⟶ F e_{(s)}, E^{\circ} = - 0.41 V$
$A g_{(a q)}^{+} + e^{-} ⟶ A g_{(s)}; E = + 0.80 V$

$E_{cell}^{\circ} = E_{cathode}^{\circ} - E_{anode}^{\circ} = + 0.80 - (- 0.41)$
$= 1.21 V$

Q. The standard oxidation potentials of iron and silver are +0.41V and −0.80V, respectively. The e.m.f. for cell reaction 2Ag(aq)+​+Fe(s)​⟶2Ag(s)​+Fe(aq)2+​ is ________ V

Reduction potential of Fe and Ag electrodes are Fe(aq )2+​+2e−⟶Fe(s)​,E∘=−0.41V Ag(aq)+​+e−⟶Ag(s)​;E=+0.80V Ecell ∘​=Ecathode ∘​−Eanode ∘​=+0.80−(−0.41) =1.21V

Q. The standard oxidation potentials of iron and silver are $+ 0.41 V$ and $- 0.80 V$ , respectively. The e.m.f. for cell reaction
$2 A g_{(a q)}^{+} + F e_{(s)} ⟶ 2 A g_{(s)} + F e_{(a q)}^{2 +}$ is ________ $V$

Reduction potential of Fe and Ag electrodes are
$F e_{(aq)}^{2 +} + 2 e^{-} ⟶ F e_{(s)}, E^{\circ} = - 0.41 V$
$A g_{(a q)}^{+} + e^{-} ⟶ A g_{(s)}; E = + 0.80 V$

$E_{cell}^{\circ} = E_{cathode}^{\circ} - E_{anode}^{\circ} = + 0.80 - (- 0.41)$
$= 1.21 V$