1928 United States presidential election in Rhode Island

1928 United States presidential election in Rhode Island

← 1924 November 6, 1928 1932 →
 
Nominee Al Smith Herbert Hoover
Party Democratic Republican
Home state New York California
Running mate Joseph Taylor Robinson Charles Curtis
Electoral vote 5 0
Popular vote 118,973 117,522
Percentage 50.16% 49.55%

County Results

President before election

Calvin Coolidge
Republican

Elected President

Herbert Hoover
Republican

The 1928 United States presidential election in Rhode Island took place on November 6, 1928, as part of the 1928 United States presidential election which was held throughout all contemporary 48 states. Voters chose five representatives, or electors to the Electoral College, who voted for president and vice president.

Rhode Island voted for the Democratic nominee, Governor Alfred E. Smith of New York, over the Republican nominee, former Secretary of Commerce Herbert Hoover of California. Smith's running mate was Senator Joseph Taylor Robinson of Arkansas, while Hoover's running mate was Senate Majority Leader Charles Curtis of Kansas.

Smith won Rhode Island by a very narrow margin of 0.61%, making him the first Democratic presidential candidate since Woodrow Wilson in 1912 to carry the state, as well as the first to win an absolute majority of the vote since Franklin Pierce in 1852 (Wilson won the state in 1912, but only with a 39.04% plurality due to Republican vote splitting between President William Howard Taft and his immediate predecessor, Theodore Roosevelt, who challenged him with a third party). Although Hoover won more counties than Smith, key to Smith's victory was his appeal to "ethnic white" Roman Catholic voters in Providence County and Bristol County. Rhode Island was the only state save adjacent Massachusetts (another state with a large Catholic population) outside the Democratic "Solid South" that voted for Smith in 1928. The former had not voted Democratic since 1912 and the latter since 1836. This was the second of three times that the state voted differently than Minnesota, along with 1912 and 1984.

Despite winning in a landslide nationally, Hoover became the first Republican to ever win the presidency without carrying Rhode Island. Given the scale of Hoover's win, Rhode Island weighed in as 18 percentage points more Democratic than the United States at large. Hoover also became the first Republican to ever win without carrying Providence and Bristol counties. Beginning in 1928, Rhode Island would transition from a strongly Yankee Republican state into a Democratic-leaning state. Since then, Republicans have only carried the state four times, all in Republican landslide years: 1952, 1956, 1972, and 1984.

Results

1928 United States presidential election in Rhode Island[1]
Party Candidate Running mate Popular vote Electoral vote
Count % Count %
Democratic Al Smith of New York Joseph Taylor Robinson of Arkansas 118,973 50.16% 5 100.00%
Republican Herbert Hoover of California Charles Curtis of Kansas 117,522 49.55% 0 0.00%
Socialist Labor Verne L. Reynolds of Michigan Jeremiah D. Crowley of New York 416 0.18% 0 0.00%
Communist William Z. Foster of Massachusetts Benjamin Gitlow of New York 283 0.12% 0 0.00%
Total 237,194 100.00% 5 100.00%

By county

1928 United States presidential election in Rhode Island (by county) [2]
County Al Smith

Democratic

Herbert Hoover

Republican

Other candidates

Various parties

Total
% # % # % # #
Bristol 51.8% 4,080 48.0% 3,780 0.2% 13 7,873
Kent 39.3% 7,460 60.4% 11,487 0.3% 58 19,005
Newport 43.9% 6,748 55.8% 8,578 0.2% 33 15,359
Providence 52.9% 97,185 46.8% 85,884 0.3% 568 183,637
Washington 30.9% 3,500 68.8% 7,793 0.2% 27 11,320

See also

References

  1. ^ "1928 Presidential General Election Results - Rhode Island". U.S. Election Atlas. Retrieved December 23, 2013.
  2. ^ "RI.gov: Election Results". www.ri.gov. Retrieved February 11, 2024.


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