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Question: Silver and copper voltmeters are connected in parallel with a battery of emf 12V. In 30min \(1g\) of...

Silver and copper voltmeters are connected in parallel with a battery of emf 12V. In 30min 1g1g of silver and 1.8g1.8g of copper are liberated. The energy supplied by the battery is
(A) 720J720J
(B) 2.41J2.41J
(C) 24.12J24.12J
(D) 4.34×104J4.34 \times {10^4}J

Explanation

Solution

We solve this problem by first finding the number of moles of silver and copper liberated. Using this we find the number of electrons required to liberate the given amount of silver and copper which is equal to the number of charges. The total number of charges is found by multiplying with Avogadro’s number. Multiplying the total number of charges with voltage we get the energy supplied.
Formula used:
Energy supplied by battery E=VitE = Vit
Where voltage is represented by VV
Current is represented by ii
Time is represented by tt

Complete step by step solution:
Number of moles of silver and copper is found by dividing the given mass with molar mass.
Molar masses of silver and copper are 108,63108,63 respectively.
Number of moles of silver in thirty minutes is 1g108\dfrac{{1g}}{{108}}
Number of moles of copper in thirty minutes is 1.8g63\dfrac{{1.8g}}{{63}}
The number of electrons required to discharge 1.8g63\dfrac{{1.8g}}{{63}} of copper and 1g108\dfrac{{1g}}{{108}} silver will be equal to the number of charges.
1108+(2×1.863)\dfrac{1}{{108}} + (2 \times \dfrac{{1.8}}{{63}})
We multiply this with Avogadro’s number to get the total charge
i=(1108+(2×1.863))×6.023×1023i = (\dfrac{1}{{108}} + (2 \times \dfrac{{1.8}}{{63}})) \times 6.023 \times {10^{23}}
Energy is equal to
E=ViE = Vi
E=12×(1108+(2×1.863))×6.023×1023E = 12 \times (\dfrac{1}{{108}} + (2 \times \dfrac{{1.8}}{{63}})) \times 6.023 \times {10^{23}}
E=4.34×104JE = 4.34 \times {10^4}J
Since we have taken the current and voltage liberated in thirty minutes, we don’t have to multiply with time again to get energy
The energy supplied is equal to E=4.34×104JE = 4.34 \times {10^4}J

Hence option (D) E=4.34×104JE = 4.34 \times {10^4}J is the correct answer

Note: Students might make a mistake of multiplying time again at the end which should not be done because all the values of time are considered.
We find the number of charges by multiplying with the number of cations or anions silver and copper form. Silver has a charge of 1 and copper has a charge of 2.