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2 years ago

Consider a voltaic cell based on the following cell reaction:

Chemistry

1 answer:

Illusion [34]2 years ago

4 0

Answer:

$E^0_{[At_2/At^-]}=0.30V$

Explanation:

$Ni+At_{2}\rightarrow Ni^{2+}+2At^-$

Here Ni undergoes oxidation by loss of electrons, thus act as anode. At undergoes reduction by gain of electrons and thus act as cathode.

$E^0=E^0_{cathode}- E^0_{anode}$

Where both $E^0$ are standard reduction potentials.

$E^0_{[Ni^{2+}/Mg]}= -0.25V$

$E^0_{[At_2/At^-]}=?$

$E^0_{cell}=0.55V$

$E^0=E^0_{[At_2/At^-]}- E^0_{[Ni^{2+}/Mg]}$

$0.55V=E^0_{[At_2/At^-]-(-0.25V)$

$E^0_{[At_2/At^-]}=0.30V$

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C2H5OH(aq) + MnO− 4 (aq) → Mn2+(aq) + CH3COOH(aq) of acetic acid from ethanol by the action of permanganate ion in acidic soluti

Andreas93 [3]

Answer :

Ethanol $(C_2H_5OH)$ act as reducing agent.

The smallest possible integer coefficient of $MnO_4^-$ in the combined balanced equation is, 4

Explanation :

Redox reaction or Oxidation-reduction reaction : It is defined as the reaction in which the oxidation and reduction reaction takes place simultaneously.

Oxidation reaction : It is defined as the reaction in which a substance looses its electrons. In this, oxidation state of an element increases. Or we can say that in oxidation, the loss of electrons takes place.

Reduction reaction : It is defined as the reaction in which a substance gains electrons. In this, oxidation state of an element decreases. Or we can say that in reduction, the gain of electrons takes place.

Reducing agent : It is defined as the agent which helps the other substance to reduce and itself gets oxidized. Thus, it will undergo oxidation reaction.

Oxidizing agent : It is defined as the agent which helps the other substance to oxidize and itself gets reduced. Thus, it will undergo reduction reaction.

The given chemical reaction is,

$C_2H_5OH(aq)+MnO_4^-(aq)\rightarrow CH_3COOH(aq)+Mn^{2+}(aq)$

The oxidation-reduction half reaction will be :

Oxidation : $C_2H_6O\rightarrow C_2H_4O_2$

Reduction : $MnO_4^-\rightarrow Mn^{2+}$

Now balance oxygen atom on both side.

Oxidation : $C_2H_6O+H_2O\rightarrow C_2H_4O_2$

Reduction : $MnO_4^-\rightarrow Mn^{2+}+4H_2O$

Now balance hydrogen atom on both side.

Oxidation : $C_2H_6O+H_2O\rightarrow C_2H_4O_2+4H^+$

Reduction : $MnO_4^-+8H^+\rightarrow Mn^{2+}+4H_2O$

Now balance the charge.

Oxidation : $C_2H_6O+H_2O\rightarrow C_2H_4O_2+4H^++4e^-$

Reduction : $MnO_4^-+8H^++5e^-\rightarrow Mn^{2+}+4H_2O$

In order to balance the electrons, we multiply the oxidation reaction by 5 and reduction reaction by 4 and then added both equation, we get the balanced redox reaction.

Oxidation : $5C_2H_6O+5H_2O\rightarrow 5C_2H_4O_2+20H^++20e^-$