1884 United States presidential election in Iowa

1884 United States presidential election in Iowa

← 1880 November 4, 1884 1888 →
 
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%

County Results

President before election

Chester A. Arthur
Republican

Elected President

Grover Cleveland
Democratic

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. State voters chose 13 electors to the Electoral College, which selected the president and vice president.[1]

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.

  1. ^ a b "1884 Presidential Election Results Iowa".

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