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3 years ago

At which electrode does oxidation occur in a

Chemistry

2 answers:

lisabon 2012 [21]3 years ago

7 0

Answer: Option (3) is the correct answer.

Explanation:

Both voltaic cell and electrochemical cells are used to produce electrical energy with the help of chemical reactions.

These cells also contain a salt bridge which helps in the neutrality of ions.

Therefore, in both the cells oxidation reaction occurs at anode and reduction reaction occurs at cathode.

Therefore, we can conclude that at the anode in both a voltaic cell and an

electrolytic cell oxidation occurs.

Charra [1.4K]3 years ago

6 0

(3) the anode in both a voltaic cell and an <span>electrolytic cell is your answer.</span>

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A. Which reactant is the limiting reagent?

Tasya [4]

Answer:

a. Zinc is the limiting reactant.

b. $m_{ZnBr_2}^{by\ Zn}=162.61gZnBr_2$

c. $m_{Br_2}^{leftover}=6.6g$

Explanation:

Hello there!

a. In this case, when zinc metal reacts with bromine, the following chemical reaction takes place:

$Zn+Br_2\rightarrow ZnBr_2$

Thus, since zinc and bromine react in a 1:1 mole ratio, we can compute their reacting moles to identify the limiting reactant:

$n_{Zn}=47.2g*\frac{1mol}{65.38g} =0.722molZn\\\\n_{Br_2}=122g*\frac{1mol}{159.8g} =0.763molBr_2$

Thus, since zinc has the fewest moles we infer it is the limiting reactant.

b. Here, we compute the grams of zinc bromide via both reactants:

$m_{ZnBr_2}^{by\ Zn}=0.722molZn*\frac{1molZnBr_3}{1molZn} *\frac{225.22gZnBr_2}{1molZnBr_2} =162.61gZnBr_2\\\\m_{ZnBr_2}^{by\ Br_2}=0.763molBr_2*\frac{1molZnBr_3}{1molBr_2} *\frac{225.22gZnBr_2}{1molZnBr_2} =171.95gZnBr_2$

That is why zinc is the limiting reactant, as it yields the fewest moles of zinc bromide product.

c. Here, since just 0.722 mol of bromine would react, we compute the corresponding mass:

$m_{Br_2}^{reacted}=0.722molBr_2*\frac{159.8gBr_2}{1molBr_2} =115.4gBr_2$

Thus, the leftover of bromine is:

$m_{Br_2}^{leftover}=122g-115.4g\\\\m_{Br_2}^{leftover}=6.6g$

Best regards!

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