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Strike441 [17]
3 years ago
5

Calculate for the following electrochemical cell at 25°C, Pt H2(g) (1.0 atm) H (0.010 M || Ag (0.020 M) Ag if E (H) - +0.000 V a

nd E(Ag)-0.799 V. a. +0.817 V b. +0.799 V c. +0.911 V d. +0.275 V e. +1.01 V ,
Chemistry
1 answer:
viva [34]3 years ago
4 0

Answer : The correct option is, (b) +0.799 V

Solution :

The values of standard reduction electrode potential of the cell are:

E^0_{[H^{+}/H_2]}=+0.00V

E^0_{[Ag^{+}/Ag]}=+0.799V

From the cell representation we conclude that, the hydrogen (H) undergoes oxidation by loss of electrons and thus act as anode. Silver (Ag) undergoes reduction by gain of electrons and thus act as cathode.

The half reaction will be:

Reaction at anode (oxidation) : H_2\rightarrow 2H^{+}+2e^-    

Reaction at cathode (reduction) : Ag^{+}+1e^-\rightarrow Ag    

The balanced cell reaction will be,  

H_2+2Ag^{+}\rightarrow 2H^{+}+2Ag

Now we have to calculate the standard electrode potential of the cell.

E^o_{cell}=E^o_{cathode}-E^o_{anode}

E^o_{cell}=E^o_{[Ag^{+}/Ag]}-E^o_{[H^{+}/H_2]}

E^o_{cell}=(+0.799V)-(+0.00V)=+0.799V

Therefore, the standard cell potential will be +0.799 V

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In one Ca(OH)₂ molecule, there are 5 atoms (1 Ca, 2 O and 2 H). This means that in 1.005x10²⁴ Ca(OH)₂ molecules, there are (5 * 1.005x10²⁴) 5.025x10²⁴ atoms.

To <u>convert molecules into moles</u>, we use <em>Avogadro's number</em>:

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What results when two waves that are completely out of phase meet?
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