1884 United States presidential election in Iowa

The 1884 United States presidential election in Iowa took place on November 4, 1884. All contemporary 38 states were part of the 1884 United States presidential election. Iowa voters chose thirteen electors to the Electoral College, which selected the president and vice president. [1]

1884 United States presidential election in Iowa

November 4, 1884
 
Nominee James G. Blaine Grover Cleveland
Party Republican Democratic
Home state Maine New York
Running mate John A. Logan Thomas A. Hendricks
Electoral vote 13 0
Popular vote 197,089 177,316
Percentage 52.25% 47.01%

President before election

Chester A. Arthur
Republican

Elected President

Grover Cleveland
Democratic

Iowa was won by Secretary of State James G. Blaine (R-Maine), running with Senator John A. Logan, with 52.25% of the vote, against Grover Cleveland, the 28th governor of New York, (DNew York), running with the former governor of Indiana Thomas A. Hendricks, with 47.01% of the popular vote.[1]

The Prohibition party chose the 8th Governor of Kansas, John St. John and Maryland State Representative William Daniel, received 0.40% of the popular vote.

Results

1884 United States presidential election in Iowa
Party Candidate Votes Percentage Electoral votes
Republican James G. Blaine 197,089 52.25% 13
Democratic Grover Cleveland 177,316 47.01% 0
Prohibition John St. John 1,499 0.40% 0
No party Write-ins 1,297 0.34% 0

References

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