1912 United States presidential election in Iowa

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1912 United States presidential election in Iowa

← 1908 November 5, 1912 1916 →
 
Nominee Woodrow Wilson Theodore Roosevelt William Howard Taft
Party Democratic Progressive Republican
Home state New Jersey New York Ohio
Running mate Thomas R. Marshall Hiram Johnson Nicholas M. Butler
(replacing James S. Sherman)
Electoral vote 13 0 0
Popular vote 185,325 161,819 119,805
Percentage 37.64% 32.87% 24.33%

President before election

William Howard Taft
Republican

Elected President

Woodrow Wilson
Democratic

The 1912 United States presidential election in Iowa took place on November 5, 1912, as part of the 1912 United States presidential election. Voters chose thirteen representatives, or electors, to the Electoral College, who voted for president and vice president.

Iowa was won by Princeton University President Woodrow Wilson (DVirginia), running with governor of Indiana Thomas R. Marshall, with 37.64% of the popular vote, against the 26th president of the United States Theodore Roosevelt (PNew York), running with governor of California Hiram Johnson, with 32.87% of the popular vote and the 27th president of the United States William Howard Taft (ROhio), running with Columbia University President Nicholas Murray Butler, with 24.33% of the popular vote.[1]

Wilson was the first Democrat to win Iowa's electoral votes since Franklin Pierce in 1852. No Democrat would obtain a majority until Franklin Roosevelt in 1932.

Results

1912 United States presidential election in Iowa[1]
Party Candidate Votes %
Democratic Woodrow Wilson 185,325 37.64%
Progressive Theodore Roosevelt 161,819 32.87%
Republican William Howard Taft 119,805 24.33%
Socialist Eugene V. Debs 16,967 3.45%
Prohibition Stuart Hamblen 8,440 1.71%
Total votes 492,356 100%

References

  1. ^ a b "1912 Presidential Election Results Iowa". {{cite web}}: Cite has empty unknown parameter: |1= (help)