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Redox Quick Practice for WA1

Total questions: 6

Worksheet time: 6mins

Name
Class
Date
1.

In the equation

Mg(s)+FeCl2(aq) → MgCl2(aq)+Fe(s)Mg(s)+FeCl_2(aq)\ \rightarrow\ MgCl_2(aq)+Fe(s)  

what is the oxidising agent? Check ALL answers that are correct

a)

Iron

b)

Fe2+Fe^{2+}  

c)

Fe

d)

FeCl2FeCl_2  

2.

In the reaction

Mg(s)+FeCl2(aq) → MgCl2(aq)+Fe(s)Mg(s)+FeCl_2(aq)\ \rightarrow\ MgCl_2(aq)+Fe(s)  

why is FeCl2 reduced, in terms of oxidation states?

a)

 The oxidation state of Fe in FeCl2FeCl_2  is +2 and it decreases to 0 in Fe

b)

The oxidation state of FeCl2FeCl_2  decreases from +2 to 0 in Fe

c)

The oxidation state of iron decreases from +2 in FeCl2FeCl_2  to 0 in Fe

d)

Iron in FeCl2FeCl_2  ​has an oxidation state of +2 and it decreases by 2 to an oxidation state of 0 in Fe

3.

Which is the correct ionic half equation for the reaction in which chlorine gas turns to chloride ions?

(Be sure to watch the explanation video after you have submitted your answer!)

a)

Cl(g) + e → Cl−(aq)Cl_{ }\left(g\right)\ +\ e\ \rightarrow\ Cl^-\left(aq\right)  

b)

Cl2(g) + 2e → 2Cl−(aq)Cl_2\left(g\right)\ +\ 2e\ \rightarrow\ 2Cl^-\left(aq\right)  

c)

Cl2(g)  → 2Cl−(aq) + 2eCl_2\left(g\right)\ \ \rightarrow\ 2Cl^-\left(aq\right)\ +\ 2e  

d)

Cl2(g) + e → 2Cl−(aq) Cl_2\left(g\right)\ +\ e\ \rightarrow\ 2Cl^-\left(aq\right)\  

e)

Cl2(g)  → 2Cl−(aq) + eCl_2\left(g\right)\ \ \rightarrow\ 2Cl^-\left(aq\right)\ +\ e  

4.

What is the ionic equation that can be obtained from the chemical equation below?

2FeCl2(aq) +Cl2(g) → 2FeCl3(aq) 2FeCl_2\left(aq\right)\ +Cl_2\left(g\right)\ \rightarrow\ 2FeCl_3\left(aq\right)\  

a)

Cl2(g) → 2Cl−(aq)Cl_2\left(g\right)\ \rightarrow\ 2Cl^-\left(aq\right)  

b)

2Fe2+(aq) → 2Fe3+(aq) 2Fe^{2+}\left(aq\right)\ \rightarrow\ 2Fe^{3+}\left(aq\right)\  

c)

2Fe2+(aq) + Cl2(g) → 2Fe3+(aq) + 2Cl−(aq)2Fe^{2+}\left(aq\right)\ +\ Cl_2\left(g\right)\ \rightarrow\ 2Fe^{3+}\left(aq\right)\ +\ 2Cl^-\left(aq\right)  

d)

Fe2+(aq) + Cl2(g) → Fe3+(aq) + Cl−(aq)Fe^{2+}\left(aq\right)\ +\ Cl_2\left(g\right)\ \rightarrow\ Fe^{3+}\left(aq\right)\ +\ Cl^-\left(aq\right)  

5.

For the following ionic half equation,

Cl2(g) + 2e → 2Cl−(aq)Cl_2\left(g\right)\ +\ 2e\ \rightarrow\ 2Cl^-\left(aq\right)  

explain why chlorine is reduced, in terms of electron transfer

a)

Cl2Cl_2  loses electrons to be reduced to Cl−Cl^-  

b)

Cl2Cl_2  gains electrons to be reduced to Cl−Cl^-  

c)

Chlorine is reduced to chloride as it loses electrons

d)

Chlorine is reduced to chloride as it gains electrons

6.

For the following ionic equation,

    (i)Cr2O72−+    (ii)H++    (iii)e → (iv)Cr3++    (v)H2O\ \ \ \ \left(i\right)Cr_2O_7^{2-}+\ \ \ \ \left(ii\right)H^++\ \ \ \ \left(iii\right)e\ \rightarrow\ \left(iv\right)Cr^{3+}+\ \ \ \ \left(v\right)H_2O  

fill in the missing coefficients (i) to (v).

Your answer should be in the form of a single 5-digit number with NO spaces and NO commas, eg 141032

if you think that the coefficients are 1, 4, 10, 3, and 2 for (i) to (v) respectively

(a)